Question

A ball is projected upward at time t = 0.0 s, from a point on a roof 20 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 37.9 m/s. Consider all quantities as positive in the upward direction. At time t = 6.2 s, the acceleration of the ball is closest to:

Answer 1

Answer:

$9.8\ m/s^2$

Explanation:

Given:

• x = initial vertical height of the ball = 20 m
• y = final height of the ball = 0 m
• a = acceleration of the ball in air = $-g = -9.8\ m/s^2$
• u = initial vertical velocity of the ball = 37.9 m/s

Assume:

• t = time interval

When an object is in air, the gravitational force due to earth in the downward direction and air resistance opposite to the motion of the object acts on it. But, air resistance is neglected unless it is mentioned in the question. This means only gravitational force due to the earth acts on it in the downward direction.

Let us first find out the time interval for which the ball remains in the air i.e., the time instant from which the ball was projected vertically up to the time instant at which the ball strikes the ground.

Since the ball in the air remains at constant acceleration motion, we have the following relation:

$y-x = ut+\dfrac{1}{2}at^2\\\Rightarrow 0-20=37.9t+\dfrac{1}{2}(-9.8)t^2\\\Rightarrow 4.9t^2-37.9t-20=0$

On solving the above quadratic equation, we have

t = -0.50 s  or   t = 8.23 s

Since t = -0.50 s is the time in the past. So, it is wrong and t = 8.23 s is the desirable time instant.

This means the ball is in the air for 8.23 s. Since the ball in the air remains in constant acceleration motion in the downward direction. So, the ball has a constant acceleration closest to $9.8\ m/s^2$ at t = 6.2 s. This is because the ball remains in the air for 8.23 s.