Question

A ball is thrown up into the air and returns to the same level. It is in the air for 4 seconds. With what initial velocity was it thrown?

Answer 1

To determine the initial velocity at which the ball was thrown, we can use the kinematic equation for vertical motion:

$h=v_{i}t-\frac{1}{2}g{t}^2$

where:

• $h$ is the maximum height reached by the ball (in this case, zero since it returns to the same level)
• $v_i$ is the initial velocity of the ball
• $t$ is the time the ball spends in the air (4 seconds)
• $g$ is the acceleration due to gravity (approximately $9.8{\text{m/s}}^2$)

Since the ball returns to the same level, the maximum height reached is zero. Plugging in the given values, the equation becomes:

$0=v_i\cdot 4-\frac{1}{2}\cdot 9.8\cdot 4^2$

Simplifying further:

0=4vi−78.40 = 4v_i - 78.40=4v<span class="pstrut" style="hei