Respuesta :
Answer:
a) 17.8 m/s
b)28.3 m
Explanation:
Given:
angle A = 53.0°, sinA = 0.8, cosA = 0.6
d = 40.0 m, h = 15.0 m, H = 100 m,
(a) find speed v
vertical displacement -h = v×sinA×t - 0.5gt^2
-15 = 0.8vt - 4.9t^2....... (1)
horizontal displacement d = v×cosA×t = 0.6×v×t
so v×t = d/0.6 = 40/0.6
plug it into (1) and get
-15 = 0.8*40/0.6 - 4.9t^2
t = 3.734 s
v = (40/0.6)/t = 17.8 m/s
(b) If his speed was only half the value found in (a), where did he land?
v = 17.8/2 = 8.9 m/s
vertical displacement = -H = v×sinA×t - gt^2/2
4.9×t^2 - 8.9×0.8×t - 100 = 0
t = 5.30 s
d = v×cosA×t = 28.3 m
(a) The speed at the top of the ramp will be 17.8 m/sec.
(b) He lands at 29.46 meters when the velocity is half of the original value.
What is speed?
Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity.
it is denoted by u for the initial speed while u for the final speed. its si unit is m/sec.
H = -15 m ( - ve shows upward motion )
g = -9.81 m/sec²
A = 53°
Velocity resolved in two direction
sinA = 0.8
cosA = 0.6
According to Newton's second law of motion
[tex]\rm {H = ut + \frac{1}{2}gt^2}[/tex]
[tex]\rm {-H = v\;sinAt - \frac{1}{2}gt^2}[/tex]
[tex]\rm {-15= v\;sin53^0 - 4.9t^2}[/tex]
[tex]\rm {-15= 0.8vt- 4.9t^2}[/tex]
Horizontal displacement
[tex]\rm{d = v_x\times t}[/tex]
[tex]\rm{d = vcos\theta\times t}[/tex]
[tex]\rm{d = v cos53^0\times t}[/tex]
[tex]\rm{d = 0.6v t}[/tex]
[tex]vt = \frac{40}{0.6}[/tex]
[tex]\rm {-15= 0.8 \times\frac{40}{0.6} - 4.9t^2}[/tex]
t = 3.74 sec.
[tex]\rm v = \frac{66.46}{3.73}\\\\\rm v = 17.81[/tex] m/sec.
[tex]\rm {y = v_0t - 4.9t^2}[/tex]
[tex]\rm {-100 = 8.9t - 4.9t^2}[/tex]
By solving the equation we get,
t = 5.5 sec
d = vcosA×t
d = 8.9 cos53°×5.5
d = 29.46 m
Hence he lands at 29.46 meters when the velocity is half of the original value.
To learn more about the speed refer to the link;
https://brainly.com/question/7359669
