An object is shot into the air with an initial velocity ( \mathrm{v}{\mathrm{o}}=25 \mathrm{~m} / \mathrm{s} ) at an angle of ( 37^{\circ} ) above the horizontal at a height of 20 m above the earth’s surface (Fig. 4 below). Take g ( =10 \mathrm{~m} / \mathrm{s}^{2} ). Find and draw the horizontal and vertical components of the projectile’s velocity in Fig. 4 (a) at the initial position, (b) at the highest point. Find (c) the maximum height ( y{\text {max }} ) to which the object rises, (d) the time for it to return to the earth, (e) the distance moved horizontally, and (f) draw the horizontal and vertical components of the velocity just before it hits the ground. (g) Draw the velocity vector just before it hits the ground and find its magnitude and direction.

Answer

Projectile Motion Step-by-Step Physics Solution

Projectile Motion – Full Step-by-Step Physics Solution

🧠 Part (a): Velocity Components

uₓ = 25 × cos(37°) = 19.97 m/s
uᵧ = 25 × sin(37°) = 15.04 m/s

🧠 Part (b): At the Highest Point

Horizontal velocity remains: 19.96 m/s
Vertical velocity becomes: 0 m/s

🧠 Part (c): Maximum Height

vᵧ² = uᵧ² − 2a(y − y₀)
0 = (15.04)² − 2 × 10 × (y − 20)
y = 31.318 m

✅ Maximum Height: 31.318 m

🧠 Part (d): Time to Return to Earth

y − y₀ = uᵧt − ½gt²
0 − 20 = 15.04t − 5t²
5t² − 15.04t − 20 = 0 → t = 4.006 s

✅ Total Time of Flight: 4.006 s

🧠 Part (e): Horizontal Distance (Range)

x = uₓ × t = 19.97 × 4.006 = 79.99 m

✅ Range: 79.99 m

🧠 Part (f): Final Velocity Components Before Impact

vᵧ = uᵧ − g × t = 15.05 − 10 × 4.006 = −25.01 m/s
vₓ = 19.97 m/s

🧠 Part (g): Final Speed and Angle

v = √(vₓ² + vᵧ²) = √(19.97² + 25.01²) = 32.03 m/s
θ = tan⁻¹(|vᵧ| / vₓ) = tan⁻¹(25.01 / 19.97) = 51.8°

✅ Final Velocity: 32.03 m/s at 51.8° below horizontal

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