Asked by Sam
                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 = ?°
            
            
        (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 = ?°
Answers
                    Answered by
            Steve
            
    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
    
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
                    Answered by
            Anonymous
            
    Can you do this using a kinematic equation approach?
    
                    Answered by
            Anonymous
            
     hmd
    
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