Sagot :
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.
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