Question

If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 14m/s, how far from the base of the dorm should Allison stand to catch the ball? Assume the vertical distance between where Emily releases the ball and Allison catches it is 4.0m.

Answer 1

Emily throws the ball at 30 degree below the horizontal

so here the speed is 14 m/s and hence we will find its horizontal and vertical components

v_x = 14 cos30 = 12.12 m/sv

x

=14cos30=12.12m/s

v_y = 14 sin30 = 7 m/sv

y

=14sin30=7m/s

vertical distance between them

\delta y = 4 mδy=4m

now we will use kinematics in order to find the time taken by the ball to reach at Allison

\delat y = v_y *t + \frac{1}{2} at^2\delaty=v

y

∗t+

2

1

at

2

here acceleration is due to gravity

a = 9.8 m/s^2a=9.8m/s

2

now we will have

4 = 7 * t + \frac{1}{2}*9.8 * t^24=7∗t+

2

1

∗9.8∗t

2

now solving above quadratic equation we have

t = 0.44 st=0.44s

now in order to find the horizontal distance where ball will fall is given as

d = v_x * td=v

x

∗t

here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time

d = 12.12 * 0.44 = 5.33 md=12.12∗0.44=5.33m