Question pattern: Horizontal projectile motion
Original question 1 of 1 in this pattern
Horizontal projectile motion
A small ball is launched horizontally at 8.0 m s⁻¹ from a platform 5.0 m above the ground. Air resistance is negligible. Take g = 9.81 m s⁻².
Calculate:
a) the time taken for the ball to reach the ground;
b) the horizontal distance travelled;
c) the speed of the ball immediately before it reaches the ground.
Worked answer
Solution
- Step 1
Use vertical motion:
5.0 = 1/2 gt². - Step 2
t = √(10 / 9.81) = 1.01 s. - Step 3
Horizontal distance:
x = 8.0 × 1.01 = 8.08 m. - Step 4
Vertical speed before impact:
vᵧ = gt = 9.90 m s⁻¹. - Step 5
Resultant speed:
v = √(8.0² + 9.90²) = 12.7 m s⁻¹.
Final answer
Time = 1.01 s; horizontal distance = 8.08 m; impact speed = 12.7 m s⁻¹.
Common mistake
Assuming gravity changes the horizontal component of velocity.
Concepts tested
- projectile motion
- vector components
- free fall
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