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Nady [450]
2 years ago
9

A ball is thrown directly downward with an initial speed of 8.75 m/s, from a height of 29.4 m. After what time interval does it

strike the ground?
Physics
1 answer:
steposvetlana [31]2 years ago
6 0

After time 1.74 seconds ball will strike the ground.

A ball is thrown directly downward and information we have-

Initial velocity of ball (u) = 8.75m/s

distance/height (s) = 29.4m

Ball is thrown downward means there are gravitation force also worked and the acceleration of ball is-

acceleration (a) = g = 9.8 m/s²

When ball strike the ground then it's velocity should be 0 means-

Final velocity of ball (v) = 0 m/s²

For getting time period we can use 2nd equation of motion as-

s = ut + (1/2)at²

Now, 29.4 = (8.75 × t) + (1/2× 9.8 × t²)

29.4 = 8.75t + 4.9t²

Now we have a quadratic equation as -

4.9t² + 8.75t - 29.4 = 0

After solving it we get two values of t -

time = 1.74 and time = -3.5

But time can not be negative so we will reject the negative values.

so t = 1.74 seconds

So to conclude that after applying the second equation of motion the time taken by the ball to reach ground is 1.74 seconds.

Learn more about free fall motion here:

brainly.com/question/12167131

#SPJ10

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