HARD · VECTORS AND PROJECTILE MOTION
VELOCITY AT GIVEN TIME
A projectile is launched from the origin at 40 m/s and 60° above horizontal. At t = 2 s after launch, find the horizontal velocity component, the vertical velocity component, the overall speed, and the angle the velocity vector makes with the horizontal.
Linked equations:v_x = v_0\cos\theta,\; v_y = v_0\sin\thetav = v_0 + at,\; d = v_0 t + \tfrac{1}{2}at^2,\; v^2 = v_0^2 + 2ad
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Step-by-step solution
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Step 1
Find the horizontal component of velocity at t = 2 s.
Hint
There is no horizontal force, so the horizontal velocity never changes. It equals v₀·cos(θ) from launch to landing.
Step 2
Find the vertical component of velocity at t = 2 s.
Step 3
What is the overall speed of the projectile at t = 2 s?
Solution walkthrough
A 60° launch at 40 m/s breaks into vₓ = 40·cos(60°) = 20 m/s horizontally and vy₀ = 40·sin(60°) ≈ 34.64 m/s vertically.
The horizontal component is locked at 20 m/s — no force acts sideways. The vertical component loses g ≈ 9.807 m/s per second: at t = 2 s, vy = 34.64 - 9.807·2 ≈ 15.03 m/s. It is still positive, meaning the ball is still climbing. (The peak, where vy = 0, arrives at t_peak = 34.64/9.807 ≈ 3.53 s.)
Speed = sqrt(20² + 15.03²) = sqrt(400 + 225.9) ≈ 25.01... wait — let us recalculate cleanly. 20² = 400, 15.03² ≈ 225.9, sum ≈ 625.9, sqrt ≈ 25.02 m/s... Actually for v₀=40, cos60°=0.5, sin60°≈0.8660: vₓ=20, vy₀≈34.64, vy(2)=34.64-19.61≈15.03, speed=sqrt(400+225.9)≈25.01 m/s. The angle is atan(15.03/20) ≈ 0.645 rad ≈ 36.9°.
Key insight: speed at any moment is less than the launch speed only because vy has decreased, not because energy has been lost to air (air is off). At the peak, the speed drops to its minimum — exactly vₓ = 20 m/s — and then rises again as the ball picks up downward velocity during the descent.
§ 02
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