Question

A football is kicked with a velocity of 32.0m/s and at an angle of 42 degrees it takes 4.40s. How high does it travel?

Answer 1

23.4 m

Explanation

to solve this we need to use the expression:

$h=v\sin (\theta)t-\frac{gt^2}{2}$

where

$\begin{gathered} h\text{ is the heigth} \\ v\text{ is initial velocity} \\ \theta\text{ is the angle} \\ t\text{ is the time} \end{gathered}$

then, let

$\begin{gathered} v=32\text{ m/s} \\ \theta=42\text{ \degree} \\ t=4.4\text{ s} \\ g=9.8\text{ }\frac{m}{s^2} \end{gathered}$

now, replace

$\begin{gathered} h=v\sin (\theta)t-\frac{gt^2}{2} \\ h=32\sin (42)(4.40)-\frac{(9.8)(4.4^2)}{2} \\ h=94.21-94.864 \\ h=-0.654 \end{gathered}$

let's check the time of flight

$\begin{gathered} t=\frac{2v\sin\theta}{g} \\ t=\frac{2\cdot32\cdot\sin(42)}{9.8} \\ t=4.36 \end{gathered}$

it means after 4.36 the ball is on the ground.

Step 2

so, to find the maximum heigth we need to use the expression

$\begin{gathered} y_{\max }=\frac{v^2\sin ^2\theta}{2g} \\ \text{replace} \\ y_{\max }=\frac{(32)^2\sin ^2(42)}{2(9.8)} \\ y_{\max }=\frac{1024\cdot0.4477}{2(9.8)} \\ y_{\max }=\frac{458.48}{19.6} \\ y_{\max }=23.39 \end{gathered}$

therefore, the answer is

23.4 m

I hope this helps you