Question
A ball is thrown straight up with an initial velocity of 19. 2 m/s (assume air resistance is negligible and take the point of release to be y0=0). Calculate the displacement of the ball (in m) after 0.5 s.
Answers
Anonymous
a = -9.81 m/s^2
v = Vi + a t = 19.2 - 9.81 * 0.5 = 14.3 m/s
h = Hi + Vi t + (1/2) a t^2 = 0 + 19.2 * 0.5 - 4.9 (0.5)^2
= 9.6 - 1.23
= 8.4 meters
v = Vi + a t = 19.2 - 9.81 * 0.5 = 14.3 m/s
h = Hi + Vi t + (1/2) a t^2 = 0 + 19.2 * 0.5 - 4.9 (0.5)^2
= 9.6 - 1.23
= 8.4 meters