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

1.7 m

Step-by-step 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+(1)/(2)gt^2\\\Rightarrow 0.7=0+(1)/(2)* 9.81t^2\\\Rightarrow t=\sqrt{(0.7* 2)/(9.81)}\\\Rightarrow t=0.38\ \text{s}`

Horizontal displacement is given by

`s_x=v_xt\\\Rightarrow s_x=4.47* 0.38\\\Rightarrow s_x=1.7\ \text{m}`

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