A projectile is fired with an initial speed of 20.0 m/s at an angle of 65 degree

ID: 1880267 • Letter: A

Question

A projectile is fired with an initial speed of 20.0 m/s at an angle of 65 degrees above the horizontal. The object hits the ground 9.50 seconds later

- How much higher or lower is the launch point relative to the point where the projectile hits the ground

-To what max height above the launch point does the projectile rise

-What is the magnitude of the projectiles velocity at the instant it hits the ground

-What is the direction (below+x) of the projectiles velocity at the instant it hits the ground

Final velocity of projectile

V= u-gt= 20 sin 65-9.8*9.5= - 74.974 m/s

As the velocity of object is negative so the launch point must above the ground

Considering motion along vertical

V^2= (u sin x) ^2 - 2gh

74.974^2= (20 sin 65)^2 +2 *9.8*h

h= 270 m

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b) height above launch point

H= (20 sin 65)^2/2*9.8= 16.763 m

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Horizontal velocity of the object

Vx= 20 cos 65= 8.452 m/s

Vertical velocity, Vy= 74.974 m/s

Net velocity

V^2= Vx^2+ Vy^2

V= 75.449 m/s

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Angle= arctan(74.974/8.452)= 83.568

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Comment in case any doubt. Goodluck

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