Question

A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00s for the ball to reach its maximum height. Find (a) the ball’s initial velocity and (b) the height it reaches.

Please check my work and if incorrect please provide detailed answer.

a)v_y=v_oy-gt
v_oy=gt
3.00s=v_o (1) 9.8m/s²
v₀=29.4m/s

b) v₀=29.4m/s; t=3.00s; a=90°
(29.4m/s)(1)(3.00s)-4.9m/s² (9s²)
88.2m-44.1m=44.1m

 

 

 

 

 

 

 

Answer 1

Concepts and reason

The concepts used to solve this problem are the kinematics equations to find the initial velocity and the maximum height.

Initially, from the final velocity expression of the linear kinematic the initial velocity of the object is calculated and from this initial velocity maximum height reach by the object is also calculated.

From the linear kinematic expression the final velocity is linearly depend on the initial velocity and the acceleration of the object.

Fundamentals

From the linear kinematic expression, the expression of the distance covered by the object in terms of the initial speed and acceleration is expresses as follows,

$s = ut + \frac{1}{2}a{t^2}$

Here, $s$ is the distance, $u$ is the initial speed, $a$ is the acceleration and $t$ is the time.

The expression of the final velocity in terms of the initial velocity and acceleration is expressed as follows,

$v = u + at$

Here, $v$ is the final velocity.

(a)

Calculate the initial velocity of the ball.

The expression of the final velocity if the object is travel upward is equal to,

$v = u - gt$

Here, $g$ is the acceleration due to the gravity.

Rearrange the above expression of the final velocity in terms of the initial velocity.

$u = v + gt$

Substitute $0$ for $v$ , $9.8 {\rm{m/}}{{\rm{s}}^2}$ for $g$ and $3.00 {\rm{s}}$ for $t$ in the above expression of the final velocity,

$\begin{array}{c}\\u = 0 + \left( {9.8 {\rm{m/}}{{\rm{s}}^2}} \right)\left( {3.00 {\rm{s}}} \right)\\\\ = 29.4 {\rm{m/s}}\\\end{array}$

Calculate the maximum height reaches by the ball.

Substitute $3.00 {\rm{s}}$ for $t$ , $9.8{\rm{ m/}}{{\rm{s}}^{\rm{2}}}$ for $g$ and $29.4 {\rm{m/s}}$ for $u$ in the above expression of the $s$

$\begin{array}{c}\\s = \left( {29.4 {\rm{m/s}}} \right)\left( {3.00 {\rm{s}}} \right) - \frac{1}{2}\left( {9.8{\rm{ m/}}{{\rm{s}}^{\rm{2}}}} \right){\left( {3.00 {\rm{s}}} \right)^2}\\\\ = 44.1 {\rm{m}}\\\end{array}$

Ans: Part a

The initial velocity of the ball is equal to $29.4 {\rm{m/s}}$ .

Part b

Therefore, the baseball reaches a height of $44.1 {\rm{m}}$ .