You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 30.5 ∘ from the ground with an initial speed of 23.5 m / s. At what height above its initial position does the softball hit the wall? Ignore any effects of air resistance.

h = 20.92 m

Explanation: By using the law of conservation of energy, the kinetic energy of the ball equals it potential energy.

Kinetic energy =mv^2/2

Potential energy = mgh

Where m = mass of the object, v = velocity of object = 23.5 m/s

g = acceleration due gravity = 9.8 m/s^2

mv^2/2 = mgh

m cancels out each other on both sides , hence we have that

v^2 = 2gh.

We want the ball to move towards the wall (horizontal motion), hence we need the horizontal component of the velocity since the velocity is inclined at an angle of 30.5 to the ground (horizontal).

Hence v = 23.5 × cos 30.5, v = 20.248 m/s

Recall that v^2 = 2gh

(20.248)^2 = 2×9.8×h

409.98 = 19.6 h

h = 409.98/ 19.6

h = 20.92 m

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