Question

A ball is thrown upward at an angle with a speed and direction such that it reaches a maximum height of 16.0 m above the point it was released, with no appreciable air resistance. At its maximum height it has a speed of 18.0 m/s. With what speed was the ball released?

Answer 1

To solve this problem it is necessary to apply the concepts related to the kinematic equations of motion for which the height of a particle is defined as a function of gravity as,

$v^2 = 2gh$

Where,

h = Height

g = Gravity constant

v = Velocity

There is not change in the horizontal component, therefore there is only the component horizontal given (18m/s) and the vertical component:

$v^2=2*g*h$

$v^2=2*9.8*16$

$v=17.70 m/s$

Applying the vector theory to find the magnitude of a vector we have to

$|\vec{V}| = \sqrt{17.7^2+18^2}$

$|\vec{V}| = 25.38m/s$

The speed at which the ball was thrown was 25.38