A soccer ball is kicked with a velocity of 12m/s, HORIZONTALLY off a building which is 8.5m high. What is the velocity of the ball after 1.25s?
This is what I found so far:
Vi = 12m/s
Dx = 15.8049336m
Dy = 8.5m
Time it took for ball to hit floor 1.31707778s
I used D=Vi(T) + 1/2a (T)^2 to find total time
I used dX = Vi(T) to find the range
Because V = D/T I decided to find Dx and Dy
Found Dy using:
D= Vi(1.25) + 1/2(9.8)1.25^2
Which gave me 7.65625m vertical distance at 1.25s
Found Dx using:
Dx = Vi(1.25) +1/2(0)1.31707778
Which gave me 15m horizontal distance at 1.25s
Because I have vertical and horizontal distance I made a triangle which would give me the velocity? Or distance?
This is what I did:
7.65625^2 + 15^2 = 16.84096684^2
Using 16.84096684m I did V = D/T and got this:
16.84096684m/1.25s = 13.47277347m/s
nope, v=d/t is average velocity. Was there something you did not understand on how I told you to do it?
Yes, if you go back to the post I don't understand the sqart or that whole equation at all, grade 11 physics! Sorry for replying late btw
To find the velocity of the ball after 1.25 seconds, you can use the Pythagorean Theorem to calculate the magnitude of the velocity. Here is the step-by-step explanation:
1. You have already determined the horizontal distance (Dx) and the vertical distance (Dy) covered by the ball after 1.25 seconds.
2. Using the equation of motion in the vertical direction, D = Vi(t) + (1/2)at^2, you found that Dy = 7.65625m.
3. Using the equation of motion in the horizontal direction, Dx = Vi(t), you found that Dx = 15m.
4. Now, using these values, you can calculate the magnitude of the displacement (D) of the ball.
D = sqrt(Dx^2 + Dy^2)
= sqrt((15m)^2 + (7.65625m)^2)
= sqrt(225m^2 + 58.70312m^2)
= sqrt(283.70312m^2)
≈ 16.84096684m (rounded to the nearest hundredth)
5. Finally, to find the velocity (V) of the ball, you can use the equation V = D/T, where T is the time taken (1.25 seconds).
V = 16.84096684m / 1.25s
≈ 13.47277m/s (rounded to the nearest hundredth)
Therefore, the velocity of the ball after 1.25 seconds is approximately 13.47277m/s.