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Soloha48 [4]
4 years ago
13

Projectile A is launched horizontally at a speed of 20 meters per second from the top of a cliff and strikes a level surface bel

ow, 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?
Physics
1 answer:
pogonyaev4 years ago
5 0

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

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The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

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When y = 2.1 m, we have;

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Solving with the aid of a graphing calculator function, we get;

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The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

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The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

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The minimum average speed the opponent must move so that he is in position to hit the ball, v_s ≈ 5.79 m/s.

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