Physics for Engineers
posted by Lauryn on .
A playground is on the flat roof of a city school, 5.2 m above the street below (see figure). The vertical wall of the building is h = 6.70 m high, forming a 1.5-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of θ = 53.0°; above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall.
a)Find the speed at which the ball clears the wall.
I calculated this correctly and got 18.13 m/s.
(b) Find the vertical distance by which the ball clears the wall.
I know that dy=vy0*t but I tried using that equation and couldn't get the answer.
I did this and it was incorrect
v0sin53 and got 18.13*sin53= 14.48
then I tried plugging that into the y, to get y=14.48sin53(2.2)-1/2(-9.8)(2.2^2)
I got 25.44-23.716=1.725, the computer assignment says that is incorrect. I don't know what I am doing wrong
(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.
I can't figure out how to do the last two, can anyone please help me set up these problems so I can figure out how to solve them? Thank you.
Check part b comments on your previous post.
thanks for your help!
You got the y component correct however you multiplied it by sin53 a second time.
The correct equation should be y=18.13sin53(2.2)-4.9(4.8)
or with the y component solved for already, y=14.48(2.2)-4.9(4.8)
The ball is at height= 8.336m
It clears the wall by 8.336-6.7=1.636m
Set the y component to 5.2m
solve the quadratic for t and plug into d=18.13cos53(t)
This will give you the horizontal distance until the ball lands on the roof.