Question

How can I calculate the maximum height of the ball? h=-16t^2+vt+h v is the initial velocity and I got 4.3cm h is the initial height and I got 14cm Range of the ball is 1.954m and the time in air is 2.1 When I solve the maximum height of the ball, it need to be ft/s.

Answer 1

Step-by-step explanation:

The path followed by the ball as a function of time is given by :

$h=-16t^2+vt+h$

Where

v is initial velocity and h is initial height,

v = 4.3 cm/s = 0.043 m/s and h = 14 cm = 0.14 m

The equation becomes,

$h=-16t^2+0.043t+0.14$ ....(1)

It is a quadratic equation,

t = 0.095 s

Put the value of t in equation (1).

$h=-16(0.095)^2+0.043(0.095)+0.14\\\\h=-0.000315\ m$

or

h = -0.00103 ft (negative sign shows height is in negative direction)