Question

You toss a ball straight up. Let t = 0 be the time just after you throw the ball, let t = treturn be the time just before it the ball returns to your hand (motion is in free fall). The only knowns are treturn, g, H (max height), Viy. Use yf = yi+Viy(deltat) -(1/2)g(delta t)^2 to determine the two times when the ball is at half of its maximum height.

Answer 1

Given that

yf = yi + Viy*(delta t) - (1/2)*g*(delta t)^2

yi = 0, since initially ball was at ground

yf = H/2, since final position at half of its maximum height

Viy = Initial vertical velocity

g = acceleration due to gravity

delta t = t - t0

t0 = time when ball is launched = 0

t = time when ball is at the half of its maximum height

H/2 = 0 + Viy*(t - 0) - (1/2)*g*(t - 0)^2

(1/2)*g*t^2 - Viy*t + H/2 = 0

Solving above quadratic equation:

t = [Viy +/- sqrt (Viy^2 - 4*(1/2)*g*(H/2))]/(2*(1/2)*g)

t = [Viy +/- sqrt (Viy^2 - g*H)]/(g)

So two times will be:

t1 = [Viy - sqrt (Viy^2 - g*H)]/g And t = [Viy + sqrt (Viy^2 - g*H)]/g

Now in your question it is mentioned that Viy is given, So in that case we can find out the two times without using treturn

But if Viy is not given than in that case:

Using 2nd kinematic equation when ball returns at landing position, then

yf = yi + Viy*t - (1/2)*g*t^2

yf = yi = 0, since ball returns at landing point, So

0 = 0 + Viy*t - (1/2)*g*t^2

t*(Viy - (1/2)*g) = 0

t = 0 & treturn = 2*Viy/g

Viy = g*treturn/2, So using this value of Viy, two times will be:

t1 = [(g*treturn/2) - sqrt ((g*treturn/2)^2 - g*H)]/g And t = [(g*treturn/2) + sqrt ((g*treturn/2)^2 - g*H)]/g

t1 = [g*treturn - sqrt (g^2*treturn^2 - 4*g*H)]/g And t = [g*treturn + sqrt (g^2*treturn^2 - 4*g*H)]/g

Let me know if you've any query.

If you need final answer in terms of treturn than use 2nd expression

If you need final answer in terms of Viy than use 1st expression