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Whitepunk [10]
3 years ago
14

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?
Physics
1 answer:
viktelen [127]3 years ago
5 0

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.

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tatiyna
Answer:  Add an incline or grade to the road track.

Explanation:
Refer to the figure shown below.

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