Question:
A ball is thrown horizontally at 20 m/s from a height of 20 m. Find its speed when it hits the ground (g = 10 m/s²).
A ball is thrown horizontally at 20 m/s from a height of 20 m. Find its speed when it hits the ground (g = 10 m/s²).
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Horizontal and vertical motions are independent; combine only at final stage using vector addition.
Updated On: Jun 20, 2026
- 28.3 m/s
- 20 m/s
- 40 m/s
- 32.6 m/s
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The Correct Option is A
Solution and Explanation
Step 1: Understanding projectile motion structure.
This is a case of horizontal projectile motion where the initial velocity is purely horizontal. The motion can be separated into independent horizontal and vertical components. Gravity affects only vertical motion, not horizontal motion.
Step 2: Horizontal motion analysis.
The horizontal velocity remains constant throughout the motion because there is no horizontal acceleration. Therefore, \( v_x = 20 \, \text{m/s} \) at all times, including at impact with the ground.
Step 3: Vertical motion setup.
Vertically, the initial velocity is zero since the ball is thrown horizontally. The only acceleration acting vertically is gravity \( g = 10 \, \text{m/s}^2 \). The ball falls freely from height 20 m under gravity.
Step 4: Finding vertical velocity at impact.
We use the kinematic equation \( v_y^2 = u_y^2 + 2gh \). Substituting values gives \( v_y^2 = 0 + 2 \times 10 \times 20 \). This simplifies to \( v_y^2 = 400 \), so \( v_y = 20 \, \text{m/s} \).
Step 5: Understanding velocity components at impact.
At the moment of hitting the ground, the velocity has two perpendicular components: horizontal \( v_x = 20 \) and vertical \( v_y = 20 \). These must be combined vectorially to get resultant velocity.
Step 6: Resultant velocity calculation.
We apply Pythagoras theorem: \( v = \sqrt{v_x^2 + v_y^2} \). Substituting values gives \( v = \sqrt{20^2 + 20^2} = \sqrt{800} \). This simplifies to \( v = 20\sqrt{2} \).
Step 7: Final numerical evaluation.
\( 20\sqrt{2} \approx 20 \times 1.414 = 28.3 \, \text{m/s} \). Hence, the final speed on impact is 28.3 m/s.
This is a case of horizontal projectile motion where the initial velocity is purely horizontal. The motion can be separated into independent horizontal and vertical components. Gravity affects only vertical motion, not horizontal motion.
Step 2: Horizontal motion analysis.
The horizontal velocity remains constant throughout the motion because there is no horizontal acceleration. Therefore, \( v_x = 20 \, \text{m/s} \) at all times, including at impact with the ground.
Step 3: Vertical motion setup.
Vertically, the initial velocity is zero since the ball is thrown horizontally. The only acceleration acting vertically is gravity \( g = 10 \, \text{m/s}^2 \). The ball falls freely from height 20 m under gravity.
Step 4: Finding vertical velocity at impact.
We use the kinematic equation \( v_y^2 = u_y^2 + 2gh \). Substituting values gives \( v_y^2 = 0 + 2 \times 10 \times 20 \). This simplifies to \( v_y^2 = 400 \), so \( v_y = 20 \, \text{m/s} \).
Step 5: Understanding velocity components at impact.
At the moment of hitting the ground, the velocity has two perpendicular components: horizontal \( v_x = 20 \) and vertical \( v_y = 20 \). These must be combined vectorially to get resultant velocity.
Step 6: Resultant velocity calculation.
We apply Pythagoras theorem: \( v = \sqrt{v_x^2 + v_y^2} \). Substituting values gives \( v = \sqrt{20^2 + 20^2} = \sqrt{800} \). This simplifies to \( v = 20\sqrt{2} \).
Step 7: Final numerical evaluation.
\( 20\sqrt{2} \approx 20 \times 1.414 = 28.3 \, \text{m/s} \). Hence, the final speed on impact is 28.3 m/s.
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