Question

You throw a baseball directly upward at time t = 0 at an initial speed of 14.7 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 = 9.80 m/s2. Maximum height? Earlier time at half maximum height? Later time at half maximum height?

Answer #1

Given,

u = 14.1 m/s ;

We know from equation of motion that

v^{2} = u^{2} + 2 a S

In our case, S = Hmax, a = -g ; and At Hmax, v = final speed = 0

.0 = (14.7)^{2} - 2 x 9.8 x Hmax

Hmax = 14.7 x 14.7/ 2 x 9.8 = 11.03 m

**Hence, Hmax = 11.03 m**

Now half of the Hmax will be, H = 11.03/2 = 5.52 m

From equation of motion

S = ut + 1/2 a t^{2}

5.52 = 14.7 x t - 1/2 x 9.8 x t^{2}

4.9 t^{2} - 14.7 t + 5.52 = 0 => t = 2.56 s and t =
0.44 s

**Hence, t = 0.44 s and t(later) = 2.56 s**

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