Question

Question 19 of 32A volleyball player bumps a ball across a net with the velocity and angleshown below. What is the maximum height of the ball?O A. 9.2 mOB. 16.3 mO C. 6.4 mOD. 13.8 m

Answer 1

The correct answer is option B. 16.3m.

The velocity, u = 19 m/s

The angle of inclination, θ = 70°

The acceleration due to gravity, g = 9.8 m/s².

The maximum height can be calculated as follows:-

`H= (u^(2) sin^(2) theta)/(2g)`

By substituting the given values in the above formula, we get:-

`H= (19^(2) (sin 70)^(2) )/(2*9.8)`

By solving this equation, we get:-

`H= (318.771)/(19.6)`

H= 16.3m.

Answer 2

Answer:

B. 16.3 m

Step-by-step explanation:

Given

The velocity, u = 19 m/s

The angle of inclination, θ = 70°

The acceleration due to gravity, g = 9.8 m/s²

The maximum height is calculated below

`H=(u^2sin^2\theta)/(2g)`

Substitute the given parameters into the formula above

`\begin{gathered} H=(19^2*(sin70)^2)/(2(9.8)) \\ \\ H=(318.771)/(19.6) \\ \\ H=16.3\text{ m} \\ \\ \end{gathered}`

The maximum height of the ball = 16.3 m