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`

Step-by-step 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 = ( u^2 sin2(\theta))/(g)`

`12.9 = ( u^2 sin2(\theta))/(g)`

Generally from trigonometric identity

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

So

`12.9 = ( 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 = (u^2 sin^2 (\theta))/(2g)`

=>
`2.70 = (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)

`(u^2 sin^2 (\theta))/(u^2 *2sin(\theta) cos(\theta)) =(52.92)/(126.42 )`

=>
`tan(\theta ) = (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 = (u sin (\theta ))/(g)`

=>
`\Delta t =  (13.322 sin (22.71 ))/(9.8)`

=>
`\Delta t = 0.5248 \ s`