MEDIUM · VECTORS AND PROJECTILE MOTION
PEAK HEIGHT AND TIME
A ball is launched at 35 m/s and 50° above horizontal from flat ground. Find the vertical component of the launch velocity, the time it takes to reach maximum height, and the peak height above the ground.
Linked equations:v_x = v_0\cos\theta,\; v_y = v_0\sin\thetaH = \frac{(v_0\sin\theta)^2}{2g}T = \frac{2v_0\sin\theta}{g}
§ 01
Step-by-step solution
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Step 1
Find the vertical component of the initial velocity.
Hint
Decompose the velocity vector: the vertical piece is v₀ times sin(θ).
Step 2
How long does it take to reach the highest point?
Step 3
What is the maximum height reached?
Solution walkthrough
Decompose the launch: the vertical component is vy = 35·sin(50°) ≈ 26.81 m/s. This is the only component that gravity can alter — the horizontal part is untouched.
Gravity decelerates the ball vertically at g ≈ 9.807 m/s². The vertical velocity starts at 26.81 m/s and drops by 9.807 m/s every second, reaching zero at t_peak = 26.81 / 9.807 ≈ 2.73 s. That is the moment the ball is at its highest point.
Peak height uses the kinematic identity: since the ball starts at vy and arrives at 0 over time t_peak, the average vertical velocity is vy/2, so H = (vy/2)·t_peak = vy²/(2g) ≈ 26.81²/19.614 ≈ 36.5 m.
The horizontal motion is irrelevant for the peak height calculation — a steeper angle with the same speed gives a higher peak (more of v₀ goes into vy), while a shallower angle keeps the ball lower but carries it further before landing. At 50°, you are tilted toward height over range, which is apparent from the 36.5 m rise versus the total range of about 122 m.
§ 02
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