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Question

Exactly 3.3 s after a projectile is fired into the air from the ground, it is observed to have a velocity v⃗ = (8.9 i^ + 4.7 j^)m/s, where the x axis is horizontal and the y axis is positive upward.

Part A Determine the horizontal range of the projectile.

Express your answer using two significant figures.

Part B Determine its maximum height above the ground

Express your answer using two significant figures.

Part C Determine the speed of motion just before the projectile strikes the ground.

Express your answer using two significant figures.

Part D Determine the angle of motion just before the projectile strikes the ground

Express your answer using two significant figures.

Answer 1

The x component of velocity (i term) is constant. The y component, v_y = v_y0 - gt. You know t and v_y (j term), so you can solve for v_y0. Once you know v_y0 you can solve for

A) R = (v₀²/g)sin(2θ₀) (Use pythagorean theorem to solve for v₀ and tan^-1(v_y0/v_x0) to solve for θ₀)

B) H = v_y0²/(2g)

C) The speed with which it will be going at ground level will be |v₀ | (It is a symmetric launch)

D) The angle will be -θ₀

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