Question

A boy standing on top of a building throws a small ball from a
height of H_{1} = 37.0 m. (See figure.) The ball leaves
with a speed of 24.5 m/s, at an angle of 53.0 degrees from the
horizontal, and lands on a building with a height of H_{2}
= 16.0 m. Calculate for how long the ball is in the air. (Neglect
air friction, and use g = 9.81 m/s^{2}.)

Answer #1

the time in the air can be found by just considering vertical
motion (y)

d_y = v_yi t + 1/2 g t^2

>>> d_y = 16.0 - 37.0 = -21.0 m (down)

>>> v_yi = v_i sin(angle) = 24.5 m/s * sin 53.0 = 19.56
m/s

>>> g = -9.81 m/s/s because I take up as positive

so

-21.0 = 19.56 t + 1/2 (-9.81) t^2

-21.0 = 19.56 t - 4.905 t^2

or

4.905 t^2 - 19.56 t - 21.0 = 0

solve this quadratic equation for t

I get 4.8 s (only the positive value makes sense in this case)

A boy standing on top of a building throws a small ball from a
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with a speed of 24.5 m/s, at an angle of 53.0 degrees from the
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air friction, and use g = 9.81 m/s2.)

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