The object is experiencing uniformly accelerated motion (UAM). We can apply the formula:
h = (vo)(t) - (1/2)(g)(t^2)
where
vo = initial velocity
t = time
g = acceleration due to gravity = 9.8 m/s^2
Solving for the height after 2 seconds,
h = 37(2) - (1/2)(9.8)(2^2)
h = 54.4 m
Then another formula to get the velocity after 2 seconds:
vf^2 - vo^2 = 2gd
where
vf = final velocity
d = distance or height
Substituting,
vf^2 - 37^2 = 2(-9.8)(54.4)
vf^2 = 302.76
vf = 17.4 m/s
By the way, at the time t = 2 s after release, the ball still doesn't reach its maximum height (if calculated, it reaches maximum height at t = 3.78 s) and at this time, velocity of ball is zero. So any time between t = 0 to 3.78 s, the velocity of ball must be less than the initial velocity.
hope this helps? `u`
Hi! i am going to apply for a medical study admission test and where just wondering if you guys could help me to come up with the formula for solving this problem. I am not looking for a answer, just the mathematical formula to solve theese kind of questions.
The question is as follows:
An object is shot vertically upward with an initial speed of 37 m/s. What is its velocity at the time of a 2 s after the shot.
1 answer