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3 years ago

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A projectile was launched horizontally with a velocity of 388 m/s, 2.89 m above the ground. How long did it take the projectile

to hit the ground? Select one: 1.34 s 0.308 s 0.768 s 0.0744 s

1 answer:

tamaranim1 [39]3 years ago

6 0

Answer:

Explanation:

Given

Velocity = 388m/s

Height S = 2.89m

Required

Time

Using the equation of motion

S =ut+1/2gt²

2.89 = 388t+1/2(9.8)t²

2.89 = 388t+4.9t²

Rearrange

4.9t²+388t-2.89 =0

Factorize

t = -388±√388²-4(4.9)(2.89)/2(4.9)

t= -388±√(388²-56.644)/9.8

t = -388±387.93/9.8

t =0.073/9.8

t = 0.00744 seconds

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