Respuesta :
The first thing we must do is write the kinematic equations for each component.
For vertical component:
Speed:
Vy = Vyi-g * t
Vyi -9.8 * t1 = 7.4
Vyi = 7.4 + 9.8 * t
Height:
y = Vyi * t- (1/2) * g * t ^ 2
7.4 * t + 9.8 * t ^ 2-4.9 * t ^ 2 = 8.9
4.9t ^ 2 + 7.4 * t-8.9 = 0
The roots are
t1 = 0.789729866757189 s
t2 = -2.299933948389842 s
Positive time, therefore:
t = 0.79 s
Then, the initial vertical velocity is:
Vyi = 7.4 + 9.8 * t
Vyi = 7.4 + 9.8 * (0.79)
Vyi = 15,142 m / s
For the maximum height we have that the final velocity is zero in the vertical component
Vy = Vyi-g * t = 0
t = Vyi / g
t = 15,142 / 9.81
t = 1.543 s
Then, the maximum height will be:
y (1,543) = (15,142) * (1,543) - (1/2) * (9.81) * (1,543) ^ 2
y = 11.68 m
The flight time is:
tv = 2 * t
tv = 2 * (1,543)
tv = 3.086 s
Then, the horizontal distance traveled is:
x = Vix * tv
x = (7.6) * (3.086)
x = 23.4536 m
After reaching the maximum height the ball falls with gravity:
11.68 = (1/2) * (9.8) * t ^ 2
Clearing t:
t = ((11.68 * 2) / (9.8)) ^ (1/2)
t = 1.543 s (Therefore, as mentioned above, tv = 2 * t)
Then,
Vyf = g * t = -9.8 * (1,543) = -15.1214
Then, Vyf = Vyi = -15.1214 m / s
Vxi = Vxf = Vx = 7.6 m / s
THE magnitude of the speed is:
v = sqrt (7.6 ^ 2 + 15.1214 ^ 2) = 16.92m / s
The angle is:
x = atan (-15.1214 / 7.6) = -63.32 deg
Answer:
y = 11.68 m
x = 23.4536 m
16.92m / s
-63.32 deg
For vertical component:
Speed:
Vy = Vyi-g * t
Vyi -9.8 * t1 = 7.4
Vyi = 7.4 + 9.8 * t
Height:
y = Vyi * t- (1/2) * g * t ^ 2
7.4 * t + 9.8 * t ^ 2-4.9 * t ^ 2 = 8.9
4.9t ^ 2 + 7.4 * t-8.9 = 0
The roots are
t1 = 0.789729866757189 s
t2 = -2.299933948389842 s
Positive time, therefore:
t = 0.79 s
Then, the initial vertical velocity is:
Vyi = 7.4 + 9.8 * t
Vyi = 7.4 + 9.8 * (0.79)
Vyi = 15,142 m / s
For the maximum height we have that the final velocity is zero in the vertical component
Vy = Vyi-g * t = 0
t = Vyi / g
t = 15,142 / 9.81
t = 1.543 s
Then, the maximum height will be:
y (1,543) = (15,142) * (1,543) - (1/2) * (9.81) * (1,543) ^ 2
y = 11.68 m
The flight time is:
tv = 2 * t
tv = 2 * (1,543)
tv = 3.086 s
Then, the horizontal distance traveled is:
x = Vix * tv
x = (7.6) * (3.086)
x = 23.4536 m
After reaching the maximum height the ball falls with gravity:
11.68 = (1/2) * (9.8) * t ^ 2
Clearing t:
t = ((11.68 * 2) / (9.8)) ^ (1/2)
t = 1.543 s (Therefore, as mentioned above, tv = 2 * t)
Then,
Vyf = g * t = -9.8 * (1,543) = -15.1214
Then, Vyf = Vyi = -15.1214 m / s
Vxi = Vxf = Vx = 7.6 m / s
THE magnitude of the speed is:
v = sqrt (7.6 ^ 2 + 15.1214 ^ 2) = 16.92m / s
The angle is:
x = atan (-15.1214 / 7.6) = -63.32 deg
Answer:
y = 11.68 m
x = 23.4536 m
16.92m / s
-63.32 deg
Let's rewrite the velocity at a height h of 8.9 m as:
[tex]v_x = 7.6~m/s[/tex]
[tex]v_{yh} = 7.4~m/s[/tex]
The motion of the ball is a parabolic motion: it is a uniform motion on x-axis (so, [tex]v_x[/tex] is constant) and an accelerated motion on the y-axis (with acceleration equal to [tex]-g[/tex], the gravitational acceleration, with a negative sign because it points downwards).
Let's solve the exercise step-by-step.
a) We can solve this part by writing the laws of motion on x and y using the data at h=8.9 m as initial data.
[tex]S_x(t)=v_x t[/tex]
[tex]S_y(t)=h+v_{yh}t- \frac{1}{2}gt^2 [/tex]
[tex]v_y(t) = v_{yh}-gt[/tex]
The maximum height [tex]h_{max}[/tex] is the height at which the velocity on the y-axis is zero. We can find the time [tex]t_{max}[/tex] at which this happens by requiring [tex]v_y(t_{max})=0[/tex]:
[tex]0=v_{yh}-gt_{max}[/tex]
from which we find
[tex]t_{max}= \frac{v_y}{g} = \frac{7.4~m/s}{9.81~m/s^2}=0.75~s [/tex]
and then substituting in [tex]S_y(t)[/tex] we find the maximum height:
[tex]S_y(t_{max})=h_{max}= 8.9~m+7.4~m/s \cdot 0.75~s-\frac{1}{2} 9.81~m/s^2 \cdot (0.75~s)^2=[/tex]
[tex]=11.7~m [/tex]
2) To find the total horizontal distance, we must find the intial conditions of the motion, at t=0. Let's call [tex]v_{y0}[/tex] the initial velocity on the y-axis (we already know the initial velocity on the x-axis, which is [tex]v_x=7.6~m/s[/tex]. Then the velocity law on the y-axis is
[tex]v_y(t) = v_{y0}-gt[/tex]
let's also call [tex]t_h[/tex] the time t at which the ball reaches the height h=8.9m. At this time [tex]t_h[/tex], [tex]v_y(t_h)=7.4~m/s[/tex]. Therefore we can write the following system of equations:
[tex]S_y(t_h) = 8.9~m = v_{y0}t_h - \frac{1}{2} 9.81 t_h^2 [/tex]
[tex]v_y(t_h) = 7.4 = v_{y0}-9.81 t_h[/tex]
Solving this system, we find
[tex]t_h=0.79~s[/tex]
and
[tex]v_{y0}=15.1~m/s[/tex]
now that we know the initial velocities on both axis, we can find the time [tex]t_0[/tex] at which the ball reaches the ground. This happens when [tex]S_y(t_0)=0[/tex]. The law of motion on y is
[tex]S_y(t)=v_{y0} t - \frac{1}{2} g t^2 [/tex]
And if we require [tex]S_y(t_0)=0[/tex], we find two solutions: t=0 (beginning of the motion) and [tex]t_0=3.1~s[/tex], which is the moment when the ball reaches the ground. To find the total horizontal distance covered we just need to put this time into the equation of [tex]S_x(t)[/tex]:
[tex]S_x(t_h)=v_x t_h = 23.4~m[/tex]
c) To solve this part of the problem, we must find the vertical velocity at the time the ball hits the ground, so:
[tex]v_y(t_h)=v_{y0}-gt_h=15.1~m/s-9.81~m/s^2\cdot3.1~s=-15.3~m/s[/tex]
which is negative because now the ball is going downwards.
So, the magnitude of the velocity just before the ball hits the ground is
[tex]v= \sqrt{v_x^2+v_y^2}= \sqrt{(7.6~m/s)^2+(-15.3~m/s)^2} = 17.1~m/s[/tex]
d) The tangent of the angle of the velocity is the ratio between the two components of the velocity:
[tex]tg \theta = \frac{v_x}{v_y} = \frac{-15.3~m/s}{7.6~m/s} [/tex]
from which we find
[tex]\theta = -63.6~^{\circ}[/tex]
[tex]v_x = 7.6~m/s[/tex]
[tex]v_{yh} = 7.4~m/s[/tex]
The motion of the ball is a parabolic motion: it is a uniform motion on x-axis (so, [tex]v_x[/tex] is constant) and an accelerated motion on the y-axis (with acceleration equal to [tex]-g[/tex], the gravitational acceleration, with a negative sign because it points downwards).
Let's solve the exercise step-by-step.
a) We can solve this part by writing the laws of motion on x and y using the data at h=8.9 m as initial data.
[tex]S_x(t)=v_x t[/tex]
[tex]S_y(t)=h+v_{yh}t- \frac{1}{2}gt^2 [/tex]
[tex]v_y(t) = v_{yh}-gt[/tex]
The maximum height [tex]h_{max}[/tex] is the height at which the velocity on the y-axis is zero. We can find the time [tex]t_{max}[/tex] at which this happens by requiring [tex]v_y(t_{max})=0[/tex]:
[tex]0=v_{yh}-gt_{max}[/tex]
from which we find
[tex]t_{max}= \frac{v_y}{g} = \frac{7.4~m/s}{9.81~m/s^2}=0.75~s [/tex]
and then substituting in [tex]S_y(t)[/tex] we find the maximum height:
[tex]S_y(t_{max})=h_{max}= 8.9~m+7.4~m/s \cdot 0.75~s-\frac{1}{2} 9.81~m/s^2 \cdot (0.75~s)^2=[/tex]
[tex]=11.7~m [/tex]
2) To find the total horizontal distance, we must find the intial conditions of the motion, at t=0. Let's call [tex]v_{y0}[/tex] the initial velocity on the y-axis (we already know the initial velocity on the x-axis, which is [tex]v_x=7.6~m/s[/tex]. Then the velocity law on the y-axis is
[tex]v_y(t) = v_{y0}-gt[/tex]
let's also call [tex]t_h[/tex] the time t at which the ball reaches the height h=8.9m. At this time [tex]t_h[/tex], [tex]v_y(t_h)=7.4~m/s[/tex]. Therefore we can write the following system of equations:
[tex]S_y(t_h) = 8.9~m = v_{y0}t_h - \frac{1}{2} 9.81 t_h^2 [/tex]
[tex]v_y(t_h) = 7.4 = v_{y0}-9.81 t_h[/tex]
Solving this system, we find
[tex]t_h=0.79~s[/tex]
and
[tex]v_{y0}=15.1~m/s[/tex]
now that we know the initial velocities on both axis, we can find the time [tex]t_0[/tex] at which the ball reaches the ground. This happens when [tex]S_y(t_0)=0[/tex]. The law of motion on y is
[tex]S_y(t)=v_{y0} t - \frac{1}{2} g t^2 [/tex]
And if we require [tex]S_y(t_0)=0[/tex], we find two solutions: t=0 (beginning of the motion) and [tex]t_0=3.1~s[/tex], which is the moment when the ball reaches the ground. To find the total horizontal distance covered we just need to put this time into the equation of [tex]S_x(t)[/tex]:
[tex]S_x(t_h)=v_x t_h = 23.4~m[/tex]
c) To solve this part of the problem, we must find the vertical velocity at the time the ball hits the ground, so:
[tex]v_y(t_h)=v_{y0}-gt_h=15.1~m/s-9.81~m/s^2\cdot3.1~s=-15.3~m/s[/tex]
which is negative because now the ball is going downwards.
So, the magnitude of the velocity just before the ball hits the ground is
[tex]v= \sqrt{v_x^2+v_y^2}= \sqrt{(7.6~m/s)^2+(-15.3~m/s)^2} = 17.1~m/s[/tex]
d) The tangent of the angle of the velocity is the ratio between the two components of the velocity:
[tex]tg \theta = \frac{v_x}{v_y} = \frac{-15.3~m/s}{7.6~m/s} [/tex]
from which we find
[tex]\theta = -63.6~^{\circ}[/tex]
