A ball is projected upward at time t= 0.0s, from a point on the roof 10m above the ground. The ball rises, then falls until it strikes the ground. The initial velocity of the ball is 58.5 m/s. At time t=5.97s, what is the approximate velocity of the ball? Neglect air resistance.
To find the approximate velocity of the ball at t=5.97s, we can use the equations of motion for the vertical motion of a projectile. Since the ball is projected upward, we need to consider the downward direction as negative.
Let's use the following variables:
- u = initial velocity of the ball = 58.5 m/s (upward)
- g = acceleration due to gravity = 9.8 m/s^2 (downward)
- h = initial height of the ball above the ground = 10 m
- t = time = 5.97 s
To find the final velocity of the ball at t=5.97s, we can use the equation:
v = u + gt
Here,
- v = final velocity of the ball at t=5.97s
Substituting the given values:
v = 58.5 m/s + (9.8 m/s^2) * (5.97 s)
Now, let's calculate the value of v:
v = 58.5 m/s + (9.8 m/s^2) * (5.97 s)
v = 58.5 m/s + 58.806 m/s
v ≈ 117.306 m/s
Therefore, the approximate velocity of the ball at t=5.97s is approximately 117.306 m/s.
Why'd you use 9.8?
Because the acceleration of gravity (g) is 9.8 m/s^2.
Oh, okay, thanks.
jip-;
Solve this equation using t = 5.97:
V = Vo - g t = 58.5 - 9.8 t
That will be zero, indicating that the ball is at its highest elevation.