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.
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`
Of course, I can help you with that! To solve this problem, you need to use the equations of motion. The equation that relates final velocity, initial velocity, acceleration, and time is:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, you already have the values for initial velocity and time, and you need to find the final velocity. The acceleration is due to the force of gravity, which is approximately 9.8 m/s².
Now, let's plug in the values into the equation:
v = u + at
v = 37 m/s + (9.8 m/s²)(2 s)
Simplifying the equation, we get:
v = 37 m/s + 19.6 m/s
v = 56.6 m/s
So, the formula to solve this problem is:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
By substituting the values into the formula and solving, you can find the final velocity at a given time.