Question

A ball thrown at an initial angle of 32 degrees and initial velocity of 15 m/s reaches a maximum height h. With what initial speed must a ball be thrown straight up to reach the same maximum height h? m/s

Answer 1

In Projectile motion maximum height is given by:

h = V0^2*(sin A)^2/(2*g)

Given that

V0 = initial velocity = 15 m/sec

A = launching angle = 32 deg

g = 9.81 m/sec^2

So,

h = 15^2*(sin 32 deg)^2/(2*9.81)

h = max height = 3.22 m

Now when ball is thrown straight upward, then to reach same max height, initial velocity should be suppose U.

At the max height speed is zero, So V = 0

Max height = 3.22 m

a = -g = -9.81 m/sec^2

So, Using 3rd kinematic equation

V^2 = U^2 + 2*a*h

U = sqrt (V^2 - 2*a*h)

Using given values:

U = sqrt (0^2 - 2*(-9.81)*3.22)

U = 7.95 m/sec = Initial speed if ball is thrown straight up

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