Question

You are watching your friend play hockey. In the course of the game, he strikes the puck in such a way that, when it is at its highest point, it just clears the surrounding 2.70 m high Plexiglas wall that is 12.9 m away.

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

    $v_y =  5.14 \  m/s$

b

$\Delta t = 0.5248 \ s$

Explanation:

From the question we are told that

The maximum height is H = 2.70

The Range is R = 12.9 m

Generally from projectile motion we have that

$Range = \frac{ u^2 sin2(\theta)}{g}$

$12.9 = \frac{ u^2 sin2(\theta)}{g}$

Generally from trigonometric identity

$sin 2(\theta) = 2sin (\theta) cos(\theta)$

So

$12.9 = \frac{ u^2 2sin(\theta) cos(\theta)}{g}$

=> $u^2 * 2sin(\theta) cos(\theta) = 12.9 * g$

$u^2 * 2sin(\theta) cos(\theta) = 12.9 * 9.8$

$u^2 *2sin(\theta) cos(\theta) = 126.42 \ \cdots (1)$

Also the maximum height is

$H = \frac{u^2 sin^2 (\theta)}{2g}$

=> $2.70 = \frac{u^2 sin^2 (\theta)}{2g}$

=> $u^2 sin^2 (\theta) = 2.70 * 2 * g$

=> $u^2 sin^2 (\theta) = 2.70 * 2 * 9.8$

=> $u^2 sin^2 (\theta) = 52.92\cdots (2)$

Dividing equation 2 by (1)

$\frac{u^2 sin^2 (\theta)}{u^2 *2sin(\theta) cos(\theta)} =\frac{52.92}{126.42 }$

=> $tan(\theta ) = \frac{52.92}{126.42 }$

=> $\theta = tan^{-1} [0.4186]$

=> $\theta =22.71^o$

So

From equation 1

$u^2 *2sin(22.71) cos(22.71) = 126.42 \ \cdots (1)$

=> $u = 13.322 \ m/s$

Generally the vertical component of the initial velocity is mathematically evaluated as

$v_y = usin (\theta)$

=> $v_y = 13.322 * sin (22.71)$

=> $v_y = 5.14 \ m/s$

Generally the time taken is mathematically represented as

$\Delta t = \frac{u sin (\theta )}{g}$

=>      $\Delta t =  \frac{13.322 sin (22.71 )}{9.8}$

=> $\Delta t = 0.5248 \ s$