Question

Emily throws a soccer ball out of her dorm window to Allison, who is waiting below to catch it. If Emily throws the ball at an angle of 30º 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 8.0m.

Answer 1

Answer is :

6.778 meters.

Explanation:

y = u t + 1/2 at²

6 m = u sinθ + 1/2 at²

6 m = (12) sin 30°t + 1/2 (9.81)t²

4.905 t² + 6 t -6 = 0

By solving, t = 0.6522 s.

The horizontal distance :

x = u cosθ t

= (12) cos 30° (0.6522)

= 6.778 meters.

Answer 2

Answer:

d = 9.1 m

Explanation:

As we know that initial speed of the ball is given as

$v_i = 14 m/s$

it is projected at an angle of 30 degree below the horizontal

so we have

$v_x = 14 cos 30$

$v_x = 12.12 m/s$

$v_y = 14 sin30$

$v_y = 7 m/s$

now we know that in y direction the displacement of the ball is 8.0 m

also the acceleration in y direction is due to gravity

so we will have

$\Delta y = v_y t + \frac{1}{2}gt^2$

$8 = 7 t + \frac{1}{2}(9.81)t^2$

$t = 0.75 s$

Now in the same time the horizontal distance moved by the ball is given as

$d = v t$

$d = 12.12 \times 0.75$

$d = 9.1 m$