Question

While in a car at 4.47 meters per second a passenger drops a ball from a height of 0.70 meters above the top of a bucket how far in front of the bucket should the passenger drop the ball such that the ball will land in the bucket?

Answer 1

Answer:

1.7 m

Explanation:

$v_x$ = Velocity of ball in x direction = 4.47 m/s

$u_y$ = Velocity of ball in y direction = 0

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

t = Time taken

$s_y$ = Vertical displacement = 0.7 m

$s_y=u_yt+\dfrac{1}{2}gt^2\\\Rightarrow 0.7=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{0.7\times 2}{9.81}}\\\Rightarrow t=0.38\ \text{s}$

Horizontal displacement is given by

$s_x=v_xt\\\Rightarrow s_x=4.47\times 0.38\\\Rightarrow s_x=1.7\ \text{m}$

The passenger should throw the ball 1.7 m in front of the bucket.