A ball is thrown horizontally off a 35 m tall building. It hits the ground 80 m from the bottom of the tower.

Question

a. what is the balls time in the air
[i got 2.67]
b. what is its initial velocity?
[i got 0]
c. What is the magnitude and direction of its velocity as it strikes the ground?

Could someone please check my answers and explain how to do question c??

Answer 1

b is wrong. should be a vector in the x dircection. will add more later

Answer 2

Well it is thrown horizontally so its initial velocity in the vertical direction is zero. That means it just plain falls while continuing at the original horizontal velocity. So:
Time to fall 35 m
0 = 35 + 0 - 4.9 t^2
t^2 = 35/4.9
t = 2.67 seconds in the air

t went 80 meters in 2.67 seconds so initial and final and in between horizontal velocity is 80/2.67 = 30 m/s

Well we know the horizontal velocity is zero during this whole experience so we need the vertical velocity when it falls for 2.67 seconds
v = a t = - 9.8 * 2.67 = - 26.2 m/s
so when it crashes it is still going 30 ahead and is going 26.2 down
magnitude = sqrt(26.2^2 +30^2) = 39.8 m/s
tangent of angle down from horizontal = 26.2/30
so angle = 41.1 degrees

Answer 3

Well it is thrown horizontally so its initial velocity in the vertical direction is zero. That means it just plain falls while continuing at the original horizontal velocity. So:
Time to fall 35 m
0 = 35 + 0 - 4.9 t^2
t^2 = 35/4.9
t = 2.67 seconds in the air

It went 80 meters in 2.67 seconds so initial and final and in between horizontal velocity is 80/2.67 = 30 m/s

Well we know the horizontal velocity is 30 m/s during this whole experience so we need the vertical velocity when it falls for 2.67 seconds
v = a t = - 9.8 * 2.67 = - 26.2 m/s
so when it crashes it is still going 30 ahead and is going 26.2 down
magnitude = sqrt(26.2^2 +30^2) = 39.8 m/s
tangent of angle down from horizontal = 26.2/30
so angle = 41.1 degrees

Answer 4

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