Question

A soccer player kicks the ball that travels a distance of 60.0 m on a level field. The ball leaves his foot at an initial speed of (v0) and an angle of 26.0° above the ground. Find the initial speed (v0) of the ball.

Answer 1

Answer:

27.3 m/s

Explanation:

We are given that

Distance travel by ball=x=60 m

$\theta=26^{\circ}$

We have to find the initial speed($v_0)$ of the ball.

$x=v_0cos\theta t$

Using the formula

$60=v_0cos 26 t$

$t=\frac{60}{v_ocos 26}=\frac{60}{v_0\times 0.899}=\frac{66.7}{v_0}$

The value of y at point of foot  of the vertical distance

y=0

$y=v_0sin\theta t-\frac{1}{2}gt^2$

Using $g=9.8m/s^2$

Using the formula

$0=v_0sin 26\times \frac{66.7}{v_0}-4.9\times (\frac{66.7}{v_0})^2$

$4.9\times \frac{(66.7)^2}{v^2_0}=0.44\times 66.7$

$v^2_0=\frac{4.9\times (66.7)^2}{0.44\times 66.7}$

$v^2_0=742.8$

$v_0=\sqrt{742.8}=27.3 m/s$

Hence, the initial speed of the ball=27.3 m/s

Answer 2

Answer:

27.3 m/s

Explanation:

Horizontal range, R = 60 m

angle of projection, θ = 26°

Let the velocity of projection is vo.

Use the formula of range of the projectile

$R = \frac{u^{2}Sin2\theta} {g}$

$60 = \frac{v_{0}^{2}Sin52}{9.8}$

vo = 27.3 m/s

Thus, the velocity of projection is 27.3 m/s.