Question

You throw a basketball with an initial upward velocity 8m/s and a horizontal velocity component 12 m/s.how much time is required for the basketball to reach at top point of trajectory

Answer 1

Step-by-step explanation:

V = u – gt.

At maximum height, V = 0.

0 = 8 – 9.8t.

t = 8 ÷ 9.8

t = 0.8163 seconds.

The maximum height is then given by:

H = ut – ½gt²

H = 8(0.8163) – ½ × 9.8(0.8163)²

H = 6.5306 – 3.2651

H = 3.266 meters.