Question

Concept Simulation 2.3 provides some background for this problem. A ball is thrown vertically upward, which is the positive direction. A little later it returns to its point of release. The ball is in the air for a total time of 7.96 s. What is its initial velocity? Neglect air resistance.

Answer 1

Answer:

39 m/s

Explanation:

Hi!

The equation of motion of the ball is:

$y(t) = y_0 + v_{0y}t-(1/2)(9.8 m/s^2)t^2$

Where y_0 is the initial horizontal position, which we will take as zero.

v_0y is the initial velocity and t is the time

To find the initial velocity we will set y(7.96s) = 0

$0 = v_{0y} (7.96 s)-(4.9 m/s^2)(7.96 s)^2\\v_{0y} =(4.9 m/s^2) (7.96 s)\\v_{0y}=39.004 m/s$