Hello, I don't understand how to do part "d" of the question. Can someone please explain it to me step by step? Do I use the 2nd equation to solve it?
Using one of the following formulas:
1. Vy = g * t + Voy
2. y = 1/2 * g * t^2 + Voy * t
3. Vy^2 = Vo^2 * g + 2 * g *t
4. (avg) V = (Voy + Vy)/2
A ball thrown down by 4 m/s from 80 meter height.
Find:
(a) How long takes to reach ground?
(b) What is velocity of ball just before hit the ground?
(c) What is velocity of ball after 3 second from thrown point?
(d) At what height from the ground velocity is 30 m/s?
I calculated the following:
(a) t = 3.65
(b) Vy = 39.8 m/s
(c) Vy = 33.4 m/s
(d) = ?
a. Correct.
b. Correct.
c. Correct.
d. h = ho-(V^2-Vo^2)/2g
h = 80-(30^2-4^2)/19.6 = 34.9 m
Or
V = Vo + gt = 30 m/s.
V = 4 + 9.8t = 30
9.8t = 26
t = 2.653 s.
h = ho - (Vo*t + 0.5g*t^2)
h = 80-4*2.653 + 4.9*2.653^2 = 34.9 m
To solve part (d) of the question, you can use the equation:
Vy^2 = Vo^2 + 2 * g * h
where Vy is the final vertical velocity, Vo is the initial vertical velocity, g is the acceleration due to gravity, and h is the height.
In this case, you are given the value of Vy (30 m/s) and you need to find the height from the ground at which the velocity is 30 m/s.
Here are the steps to solve for the height:
Step 1: Rearrange the equation to solve for h:
h = (Vy^2 - Vo^2) / (2 * g)
Step 2: Substitute the known values into the equation:
h = (30^2 - 4^2) / (2 * 9.8)
Step 3: Evaluate the equation:
h = (900 - 16) / 19.6
h = 884 / 19.6
h ≈ 45.10 meters
Therefore, at a height of approximately 45.10 meters from the ground, the velocity of the ball is 30 m/s.