Question

A player throws a ball upwards with an initial velocity of 29.4 m/s. Find the maximum height it attains.
A. O
- 44.1 m
B. O
1.5 m
C. O
444
D. O
44.1 m

Answer 1

Answer:

44.1 m

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 29.4 m/s

Here,

• Final velocity (v) will be 0.
• Acceleration due to gravity (g) will be –9.8 m/s. (As it is thrown in upward direction.)

We are asked to calculate the maximum height it attains.

By using the third equation of motion for fréély falling bodies,

$\\ \longrightarrow \sf{\quad { v^2 - u^2 = 2gh }}$

• v denotes final velocity
• u denotes initial velocity
• g denotes acceleration due to gravity
• h denotes height

Substituting values,

$\\ \longrightarrow \sf{\quad { (0)^2 - (29.4)^2 = 2 \times (-9.8) \times h }}$

$\\ \longrightarrow \sf{\quad { 0 - 864.36 = -19.6 \times h }}$

$\\ \longrightarrow \sf{\quad { - 864.36 = -19.6 \times h }}$

$\\ \longrightarrow \sf{\quad {\cancel{ \dfrac{- 864.36}{-19.6} }= h }}$

$\\ \longrightarrow \quad \underline{\boxed { \textbf{\textsf{44.1 \; m = h}}} }$

Therefore, the maximum height it attains is 44.1 m.

$\underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$

More Information :

Equations of motion for fréély falling bodies :

$\boxed{ \begin{array}{cc} \pmb{\sf{ \quad \: v = u + gt \quad}} \\ \\ \pmb{\sf{ \quad \: h= ut + \cfrac{1}{2}g{t}^{2} \quad \: } } \\ \\ \pmb{\sf{ \quad \: {v}^{2} - {u}^{2} = 2gh \quad \:}}\end{array}}$

• v denotes final velocity
• u denotes initial velocity
• g denotes acceleration due to gravity
• t denotes time
• h denotes height