Question #330206

A man standing on the roof of a building throws a ball with a velocity 18.0 ms-1 at an angle 50o from the horizontal. The height of the building is 35.0 m. (Given g = -9.81 ms-2) Calculate:



(i) the maximum height of the ball from the ground,



(ii) the magnitude of the velocity of the ball just before it strikes the ground.

1
Expert's answer
2022-04-21T13:02:33-0400

1)

H=h+v2sin2α2g=44.7 m.H=h+\frac{v^2\sin^2\alpha}{2g}=44.7~m.

2)

vx=vcosα=11.57 ms,v_x'=v\cos\alpha=11.57~\frac ms,

vy=2gH=29.6 ms,v_y'=\sqrt{2gH}=29.6~\frac ms,

v=vx2+vy2=31.78 ms.v'=\sqrt{v_x'^2+v_y'^2}=31.78~\frac ms.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS