Respuesta :
v ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t²
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m
GOOD LUCK AND HOPE IT HELPS U
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t²
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m
GOOD LUCK AND HOPE IT HELPS U
The difference in horizontal distance of the ball is required.
The ball would have landed 2.5 m farther.
h = Height of spike = 2.1 m
u = Initial velocity = 17 m/s
[tex]\theta[/tex] = Angle = [tex]15^{\circ}[/tex]
Displacement in y direction is given by
[tex]h=u_yt_1+\dfrac{1}{2}a_yt_1^2\\\Rightarrow 2.1=17\sin 15t_1+\dfrac{1}{2}\times 9.81t_1^2\\\Rightarrow 4.905t_1^2+4.4t_1-2.1=0\\\Rightarrow 4905t_1^2+4400t_1-2100=0[/tex]
Solving the equation
[tex]t_1=\frac{-4400\pm \sqrt{4400^2-4\times4905\left(-2100\right)}}{2\times 4905}\\\Rightarrow t_1=0.34,-1.24[/tex]
The time taken by the ball to reach the ground is 0.34 s.
Displacement is given by
[tex]x_1=u_xt_1\\\Rightarrow x_1=17\cos 15\times 0.34\\\Rightarrow x_1=5.6\ \text{m}[/tex]
[tex]\theta=7^{\circ}[/tex]
Displacement in y direction is given by
[tex]h=u_yt_2+\dfrac{1}{2}a_yt_2^2\\\Rightarrow 2.1=17\sin 7t_2+\dfrac{1}{2}\times 9.81t_2^2\\\Rightarrow 4.905t_2^2+2.07t_2-2.1=0\\\Rightarrow 4905t_2^2+2070t_2-2100=0[/tex]
Solving the equation
[tex]t_2=\frac{-2070\pm \sqrt{2070^2-4\times4905\left(-2100\right)}}{2\times4905}\\\Rightarrow t_2=0.48,-0.89[/tex]
Displacement in x direction
[tex]x_2=u_xt_2\\\Rightarrow x_2=17\cos 7\times 0.48\\\Rightarrow x_2=8.1\ \text{m}[/tex]
Difference in distance is
[tex]\Delta x=x_2-x_1\\\Rightarrow \Delta x=8.1-5.6\\\Rightarrow \Delta x=2.5\ \text{m}[/tex]
The ball would have landed 2.5 m farther.
Learn more:
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