A ball is thrown straight up into the air with a velocity of +20.5 m/s. Find the velocity of the ball after 2.5 s.
v = Vi - g t
v = 20.5 - 9.8 (2.5)
v = 20.5 - 24.5 = -4.0
Of course if it started at the ground, the final velocity is zero because it hit the ground :)
Well, let me do some quick calculations here. If the ball is thrown straight up with a velocity of +20.5 m/s, and we neglect air resistance, then we can assume that the acceleration due to gravity is -9.8 m/s².
Now, after 2.5 seconds, let's see how high the ball has gone first. We can use the kinematic equation:
h = h0 + v0t + (1/2)at²,
where h is the height, h0 is the initial height (which we'll assume to be zero since the ball is thrown straight up), v0 is the initial velocity, t is the time, and a is the acceleration.
Plugging in the values, we have:
h = 0 + (20.5 m/s)(2.5 s) + (1/2)(-9.8 m/s²)(2.5 s)².
Solving this equation, we get h = 25.625 meters. Therefore, the ball has reached a height of 25.625 meters after 2.5 seconds.
Now, to find the velocity of the ball after 2.5 seconds, we can use the equation:
v = v0 + at,
where v is the final velocity.
Plugging in the values, we have:
v = 20.5 m/s + (-9.8 m/s²)(2.5 s).
Solving this equation, we get v = -2.5 m/s.
So, after 2.5 seconds, the velocity of the ball is -2.5 m/s. Now, you might ask why it got a negative sign. Well, that's because the ball is moving downward at this point. I guess gravity has a way of bringing things back down to earth, even when it comes to velocities. Hope that answers your question!
To solve this problem, you can use the kinematic equation:
v = u + at
Where:
v = final velocity of the ball
u = initial velocity of the ball
a = acceleration of the ball (which is equal to the acceleration due to gravity, -9.8 m/s^2)
t = time
Given:
u = +20.5 m/s (upward direction)
t = 2.5 s
Now, substitute the given values into the equation:
v = u + at
v = 20.5 m/s + (-9.8 m/s^2)(2.5 s)
To find the value of v, we need to evaluate the right side of the equation:
v = 20.5 m/s - 24.5 m/s
v ≈ -4 m/s
Therefore, the velocity of the ball after 2.5 s is approximately -4 m/s.
To find the velocity of the ball after 2.5 seconds, we can use the kinematic equation for vertical motion:
v = u + gt
Where:
v = final velocity
u = initial velocity
g = acceleration due to gravity (approximately -9.8 m/s² for objects moving upward)
t = time
In this case, the ball is thrown straight up, so the initial velocity (u) is +20.5 m/s. The acceleration due to gravity (g) is -9.8 m/s² as the ball is moving upward against gravity. The time (t) is 2.5 seconds.
Plugging in the given values into the equation, we have:
v = (+20.5 m/s) + (-9.8 m/s²) * (2.5 s)
Now we solve for v:
v = 20.5 m/s - 24.5 m/s
v = -4 m/s
The velocity of the ball after 2.5 seconds is -4 m/s. The negative sign indicates that the ball is moving downward.