Rearranging formulas is all about simple algebra rules. Just like when solving for x in an equation, you're just isolating whichever variable you want. I'll work this one out for you and hopefully it'll help, but if you need more explanation, then feel free to comment!
D = ViT + 0.5at² Subtract ViT from both sides
D - ViT = 0.5at² Divide both sides by 0.5t²
I wrote this step out a little more to show how your fraction will cancel
= a I like to flip these around so the single variable is on the right
a =
A. is meaningless.
B. is perigee.
C. is apogee.
D. is perihelion.
Answer:
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Explanation:
Hi there!
The equations of height and velocity of the ball are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity of the ball at time t.
Placing the origin at the throwing point, y0 = 0.
Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.
v = v0 + g · t
6.00 m/s = 12.0 m/s -9.81 m/s² · t
(6.00 - 12.0)m/s / -9.81 m/s² = t
t = 0.612 s
Now, let´s calculate the height of the baseball at that time:
y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)
y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²
y = 5.51 m
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Have a nice day!
