Every second, the velocity decreases by 9.8 until the velocity reaches zero. (3 seconds) It will take three seconds to go back down too!
A ball is thrown in the air with an initial velocity of 29.4 m/s upward.
a) How long does it take the ball to go up?
b) come down after reaching maximum height?
c) What is the accelleration at the top?
The simple anser is use this equation d=from the big six:
vf = vi + a t
The acceleration due to gravity is -9.8 m/s2 . Always. Even at the top. a= -9.8
The velocity of the ball at the top is zero. vf = 0
The initial velocity is 29.4 m/s. vi= 29.4
Using math we solve for t = 3 sec. (0= 29.4 + 9.8 t)
Or we can think of it visually:
Every second, the velocity decreases by 9.8 until the velocity
reaches zero. (3 seconds) It will take three seconds to go back down
too!
Answers a) 3 sec b) 3 sec c) 9.8 m/s2