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 2.96 m high Plexiglas wall that is 14.5 m away. (Ignore any effects due to air resistance.)

(a) Find the vertical component of its initial velocity.
v0y = ?m/s

(b) Find the time it takes to reach the wall.
Δt? = s

(c) Find the horizontal component of its initial velocity, and its initial speed and angle.
v0x =? m/s
v0 = ?m/s
θ = ?°

You know that the max height is reached at v^2/2g sin^2θ, and the range is v^2/g sin2θ

So,

v^2/19.6 sin^2 θ = 2.96
v^2/4.9 sinθ cosθ = 29
Dividing that, we have
tanθ = 0.408
θ = 22.2°
v = 20.15 m/s

So, the y-component of v is v sinθ = 7.617 m/s
The x-component is 18.66 m/s

That makes the time to travel 14.5 meters 0.777 seconds

Can you do this using a kinematic equation approach?

hmd

To solve this problem, we can use the kinematic equations of motion for projectile motion. We'll start by breaking down the given information and the unknowns:

Given:
- Height of the Plexiglas wall (h) = 2.96 m
- Distance to the wall (d) = 14.5 m
- The projectile just clears the wall at its highest point

Unknowns:
(a) Vertical component of initial velocity (v0y)
(b) Time taken to reach the wall (Δt)
(c) Horizontal component of initial velocity (v0x), initial speed (v0), and launch angle (θ)

Let's solve for each unknown one by one.

(a) Vertical component of initial velocity (v0y):
At the highest point of the projectile's path, the vertical velocity component will be zero. Using the equation for vertical displacement (Δy) during projectile motion:

Δy = v0y * t - (0.5 * g * t^2)

Considering the highest point, Δy = h = 2.96 m, and the acceleration due to gravity (g) = 9.8 m/s^2. We also know that the time (t) taken to reach the highest point is half of the total time of flight. Thus:

t = Δt/2

Substituting these values and rearranging the equation, we can solve for v0y:

h = v0y * (Δt/2) - (0.5 * g * (Δt/2)^2

Simplifying the equation, we get:

v0y = (h + 0.5 * g * (Δt/2)^2) / (Δt/2)

Note that we need to calculate Δt first to find the vertical component of the initial velocity.

(b) Time taken to reach the wall (Δt):
To find the time taken to reach the wall, we can use the horizontal distance (d) and horizontal component of initial velocity (v0x). We know that:

d = v0x * Δt

Rearranging the equation, we can solve for Δt:

Δt = d / v0x

(c) Horizontal component of initial velocity (v0x), initial speed (v0), and launch angle (θ):
Using the equation for horizontal distance (d) traveled during projectile motion, we have:

d = v0x * Δt

Since we already have the value of d and Δt, we can solve the equation for v0x:

v0x = d / Δt

Once we have v0x, we can use it to find the initial speed (v0) using the Pythagorean theorem:

v0 = √(v0x^2 + v0y^2)

And finally, to find the launch angle (θ), we can use the tangent of the launch angle equation:

θ = tan^(-1)(v0y / v0x)

That covers the steps needed to solve the problem. Now, we can substitute the numbers provided in the problem to calculate the answers.