Maximum height Earlier time at half maximum height Waiter t

You throw a baseball directly upward at time t-0 at an initial speed of 13.9 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 980 ms Maximum height Number Eartier time at half maximum height: Number The Later time at half maximum height: O He o wel o Tech careers partnersprivacy pol

Solution

In vertical,

initial velocity, u = 13.9 m/s

acc; , a = - 9.80 m/s^2

at maximum height, velocity will be zero so v = 0

Applying, v^2 - u^2 = 2 a d

0^2 - 13.9^2 = 2(-9.8) (H)

H = 9.86 m .............Ans

for half the maximum height, displacement d = H/2 = 4.93 m

Applying, d = u t + a t^2 /2

4.93 = 13.9 t - 9.80 t^2 /2

4.9 t^2 - 13.9 t + 4.93 = 0

t = 0.416 sec OR 2.42 sec

Earlier time = 0.416 s

Later time = 2.42 s