A ball is thrown downward from the top of a 10-meter tall building with an initial speed of 4m/s. How long does it take for the ball to hit the ground below?

Answer 1

Distance is the product of average velocity and time:

d = v * t

In this case, the velocity starts at 4 m/s and then increases due to the acceleration of gravity. The equation is v = 4 + 9.8t 9.8 is the acceleration of gravity in m/s²

The initial velocity is 4, and the final is 4 + 9.8t, so the average is (8 + 9.8t) / 2 =

4 + 4.9t

So, d = (4 + 4.9t)*t = 4.9t² + 4t

Set this equal to 10 and solve for t:

4.9t² + 4t = 10

4.9t² + 4t - 10 = 0

Does not look easy to factor, so use the quadratic formula:

t = ( -4 ±√(4² - 4*4.9*(-10)) ) / 2*4.9

t = ( -4 ± 14.56 ) / 2*4.9

discard the negative answer so:

t = 10.56 / 9.8 = 1.078 sec

CHECK: final velocity is 4 + 1.078*9.8 = 14.56

average velocity is 18.56 / 2 = 9.28

9.28 * 1.078 = 10