A ball rolls over the edge of a table with a horizontal velocity v m/s. The height of the table is 1.6 m and the horizontal range of the ball from the base of the table is 20 m. How long does it take for the ball to hit the ground?

Answer :

Answer:

0.57 seconds

Explanation:

If we just want to know how long it takes for the ball to hit the ground, it means only things happening in the vertical direction matter. The the horizontal velocity v and the 20m range of the ball don't matter for this question.

Given:

d = 1.6m

a = 9.8 m/s^2

vi = 0 m/s (this is the initial VERTICAL speed)

t = ?

d = vi * t + 0.5at^2

1.6 = 0 + 0.5(9.8)t^2

1.6 = 4.9t^2

t^2 = 0.33

t = 0.57

The time taken to hit the ground by the ball is required.

The time taken to hit the ground is 0.57 s.

v = Horizontal velocity

y = Height of table = 1.6 m

x = Displacement in x direction = 20 m

From the kinematic equations of projectile motion we have

[tex]y=\dfrac{1}{2}at^2\\\Rightarrow t=\sqrt{\dfrac{2y}{a}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1.6}{9.81}}\\\Rightarrow t=0.57\ \text{s}[/tex]

The time taken to hit the ground is 0.57 s.

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