Question

A ball is thrown upward at an angle with a speed and direction such that it reaches a maximum height of 19.5 m above the point it was released, with no appreciable air resistance. At its maximum height it has a speed of 18.5 m/s. With what speed was the ball released?a. 26.9 m/sb. 32.3 m/sc. 37.7 m/sd. 16.2 m/se. 21.5 m/s

Answer 1

Answer:

26.91 m/s

Explanation:

at maximum height it has a speed of 18.5 m/s which must be its horizontal component of the speed with which it was projected

Vₓ = 18.5 m/s

Vy = √ 2gh = √ (2 × 9.8 × 19.5 ) = 19.55 m/s

V ( velocity with which it was projected ) =√ ( 18.5² + 19.55²) = 26.91 m/s

Answer 2

Answer:

a. 26.9 m/s

Explanation:

Parameters given:

Height reached by ball, s = 19.5 m

Speed at maximum height (final velocity), v = 18.5 m/s

To find the initial speed of the ball, we use one of Newton's equations of motion:

$v^2 = u^2 - 2gs$

where g = acceleration due to gravity

The negative sign is present because the ball is thrown upward, hence, it moves against the gravitational force.

Therefore:

$18.5^2 = u^2 - (2*9.8*19.5)\\\\\\u^2 = 18.5^2 + 382.2 \\\\\\u^2 = 724.4\\\\\\u = \sqrt{724.4} = 26.9 m/s$

The ball was released at a speed of 26.9 m/s.