Question

A soccer player takes a free kick from a spot that is 17 m from the goal. The ball leaves his foot at an angle of 38 ∘, and it eventually hits the crossbar of the goal, which is 2.4 m from the ground.Question: At what speed did the ball leave his foot?

Answer 1

Answer:

The ball will leave the player's foot at 17.8 m/s

Explanation:

x = v0 [email protected] t

y = v0 [email protected] t - 1/2 g t^2

eliminating t on get the trajectory

y = x [email protected] - g x^2 / [2(v0 [email protected])^2]

then

v0 = (x / [email protected]) [g / [2(x [email protected] - y)]]^1/2

where

x = 29 m

y = 2.4 m

@ = 41 degrees

then

v0 = (29 / cos41) * [9.8 / [2*(29 * tan41 - 2.4)]]^1/2 = 17.81 m/s

Answer 2

Answer:

$v=14.5(m/s)$.

Explanation:

This is a projectile motion problem, so for solving it, we will require some equations of uniform motion and free-fall.

First, we need to recognize the data that we have so we can start solving the problem.

For Free -Fall:

$g= -9.8(m/s^{2})$

$\Delta y=2.4(m)$

For uniform motion:

$\Delta x=17(m)$

The time that takes to the ball to travel 17m horizontally and to hit the crossbar at 2.4m of height is the same, so time is a common variable for free-fall and uniform motion.

Using the equations:

$\Delta y=v_{y0}t+0.5gt^{2}$ and

$\Delta x=v_{x}t$

and noticing that

$v_{y0}=v*sin(38)$ and

$v_{x}=v*cos(38)$,

we obtain

$\Delta y=v*sin(38)t+0.5gt^{2}$ and

$\Delta x=v*cos(38)t$

wich is a system of two equations and to variables that can be easily solve.

Making

$v=\frac{\Delta x}{t*cos(38)}$

we get

$\Delta y=(\frac{\Delta x}{t*cos(38)})sin(38)t+0.5gt^{2}$,

$\Delta y=\Delta x*tan(38)+0.5gt^{2}$,

$\Delta y-\Delta x*tan(38)=0.5gt^{2}$,

$t=\sqrt{\frac{\Delta y-\Delta x*tan(38)}{0.5g}}$,

$[tex]t=\sqrt{\frac{2.4-17*tan(38)}{0.5*(-9.8)}}$

so

$t=1.50s$,

now the speed can be easily compute from one of our equations. Using

$v=\frac{\Delta x}{t*cos(38)}$,

$v=\frac{17}{1.5*cos(38)}$,

$v=14.5(m/s)$.