Question

Projectile A is launched horizontally at a speed of 20 meters per second from the top of a cliff and strikes a level surface below, 3.0 seconds later. Projectile B is launched horizontally from the same location at a speed of 30 meters per second. How long does it take Projectile B to reach the ground?

Answer 1

Answer: 3 seconds

Explanation:

Initial velocity(u) of projectile A in vertical direction = 0m/s

acceleration due to gravity a=g=9.81m/s^2

Time taken(t) of projectile A = 3s

Initial velocity of projectile B = 0m/s(vertical direction)

We can get height of cliff using parameters of projectile A since it's the same location.

Height(S) = u×t + 0.5×a×t^2

u =0

S= 0.5×9.81×3^2 = 44.145m

Time taken for projectile B to reach the ground:

S = u×t + 0.5×a×t^2

u =0, S=44.145m, a=9.81m/s^2

44.145 = 0.5×9.81×t^2

44.145 = 4.905×t^2

44.145 ÷ 4.905 = t^2

9 = t^2

t = sqrt(9)

t = 3seconds

Answer 2

The time taken for Projectile B to reach the ground is 3 seconds.

The following is necessary to solve the question:

• Initial velocity = 0 m/s
• Acceleration due to gravity = 9.81 m/s²
• Time taken for projectile A = 3s

• Initial velocity of projectile B = 0 m/s

The height will be:

= 0.5 × a × t²

= 0.5 × 9.81 × 3²

= 44.145m.

The time taken will then be calculated thus:

44.145 = 0.5 × 9.81 × t²

t² = 44.145 / (0.5 × 9.81)

t² = 44.145 / 4.905

t² = 9

t = ✓9

t = 3

Therefore, the time taken for Projectile B to reach the ground is 3 seconds.

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