Motion in a Plane: Projectile Hits Ground 7.5s Later

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SUMMARY

The projectile is launched with an initial speed of 30 m/s at a 60° angle and impacts the ground after 7.5 seconds. The vertical component of the initial velocity is calculated as 30 sin 60, while the final vertical velocity is determined using the equation v = u + at, resulting in a downward velocity of 47.52 m/s. The angle of impact is found to be 72.48° below the horizontal, leading to a relative launch angle of 12.45° above the horizontal. The vertical displacement at impact is confirmed to be zero, indicating that the launch point and impact point are at the same vertical level.

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Homework Statement



A projectile is fired with an initial speed of 30m/s at an angle of 60° above the horizontal.The object hits the ground 7.5s later.
a)How much higher or lower is the launch point relative to the point where the projectile hits the ground?

Homework Equations





The Attempt at a Solution


a)u(y)=30sin60,u(x)=30cos60
v=u+at
=30sin60-9.8*7.5
=47.52 downwards
tanθ=V(y)/V(x)
=(-47.53/30cos60)
=72.48° below the x-axis
therefore,the launch point is 72.48°-60°=12.45° above relative to the point where the projectile hits the ground.
 
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The vertical displacement, y, is the point where the projectile hits the ground.
 
Isn't the vertical displacement zero?
 

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