Physics help again!
posted by Hayden .
A ball is thrown horizontally off a 35 m tall building. It hits the ground 80 m from the bottom of the tower.
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??

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

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 
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 
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