Respuesta :
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:
[tex]v^2 = u^2 - 2gs[/tex]
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:
[tex]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[/tex]
The ball was released at a speed of 26.9 m/s.
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
