Respuesta :
Answer:
(A). The maximum height above the roof reached by the rock is 16.48 m.
(B). The velocity of the rock before hitting the ground is 37.6 m/s.
(C). The horizontal distance from the base of the building to the point where the rock strikes the ground is 122.7 m.
Explanation:
Given that,
Height = 17.0 m
Velocity = 33.0 m/s
Angle = 33.0°
(A). We need to calculate the maximum height above the roof reached by the rock
Using formula of height
[tex]h=\dfrac{v^2}{2g}[/tex]
[tex]h=\dfrac{(v_{0}\sin\theta)^2}{2g}[/tex]
Put the value into the formula
[tex]h=\dfrac{(33\times\sin33)^2}{2\times9.8}[/tex]
[tex]h=16.48\ m[/tex]
(B). We need to calculate the time
Using equation of motion
[tex]-h=v_{y}t-\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]-17.0=(v_{0}\sin\theta)t-4.9t^2[/tex]
[tex]-17.0=(33.0\times\sin33)t-4.9t^2[/tex]
[tex]4.9t^2-17.9t-17.0[/tex]
[tex]t = 4.435\ sec[/tex]
We need to calculate the magnitude of the velocity of the rock just before it strikes the ground
Using equation of motion
[tex]v_{y}=v_{0}\sin\theta-gt[/tex]
[tex]v_{y}=(33.0\times\sin33)-9.8\times4.435[/tex]
[tex]v_{y}=-25.48\ m/s[/tex]
The velocity of the rock before hitting the ground is
[tex]v=\sqrt{v_{x}^2+v_{y}^2}[/tex]
[tex]v=\sqrt{(v_{0}\cos\theta)^2+v_{y}^2}[/tex]
[tex]v=\sqrt{(33\cos33)^2+(-25.48)^2}[/tex]
[tex]v=37.6\ m/s[/tex]
(C). We need to calculate the horizontal distance from the base of the building to the point where the rock strikes the ground
Using equation of motion
[tex]R=v_{x}\times t[/tex]
[tex]R=v_{0}\cos\theta\times t[/tex]
Put the value into the formula
[tex]R=33\times\cos33\times4.435[/tex]
[tex]R=122.7\ m[/tex]
Hence, (A). The maximum height above the roof reached by the rock is 16.48 m.
(B). The velocity of the rock before hitting the ground is 37.6 m/s.
(C). The horizontal distance from the base of the building to the point where the rock strikes the ground is 122.7 m.
