# Question ec3a7

Jul 8, 2015

Maximum height: 20 m
Initial velocity: 20 m/s

#### Explanation:

When an object is thrown straight up, meaning that its angle of launch forms a ${90}^{\circ}$ angle with the horizontal, the time it needs to reach maximum height will be equal to the time it needs to return on the ground.

${t}_{\text{up" = t_"down}}$

Since you know tha it takes 4 seconds for the ball to return to the thrower, it follows that

${t}_{\text{up" = t_"down" = "2 s}}$

To determine the initial velocity of the ball, use the fact that, at maximum height, the velocity of the ball is equal to zero.

underbrace(v_"max h")_(color(blue)("=0")) = v_0 - g * t_"up"

This implies that the ball was thrown with an initial velocity of

v_0 = g * t_"up" = 10"m"/"s"^cancel(2) * 2cancel("s") = color(green)("20 m/s")

To get the maximum height of the ball, use the equation that describes its vertical displacement

${h}_{\text{max" = v_0 * t_"up" - 1/2 * g * t_"up}}^{2}$

h_"max" = 20"m"/cancel("s") * 2cancel("s") - 1/2 * 10"m"/cancel("s"^2) * 2^2cancel("s"^2) = color(green)("20 m")#