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
Thank you!
vertical velociy= 16t
so figure that at t=1.25sec
then
velocity=sqart(12^2 + vertVel^2)
and you have it
Now if you want to find the direction of the velocity, you have to figure that, I assume you just want speed.
I am sorry I do not know what sqart is, and also sorry for not replying sooner bobpursley, but I tried to solve it myself and this is what I did, perhaps I used sqart?
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
I did this before you gave me my answer but appreciate it anyways, can you tell me if I did it correctly? Im in grade 11 physics and have never heard of the term sqart so im not sure exactly if I did what you did.
To find the velocity of the ball after 1.25 seconds, we need to calculate its horizontal and vertical components separately.
First, let's find the horizontal component (Vx) of the velocity. We know that the horizontal velocity remains constant throughout the motion because there are no horizontal forces acting on the ball. Therefore, Vx remains at 12 m/s throughout.
Now, let's find the vertical component (Vy) of the velocity. We can use the equation:
Dy = Viy * T + (1/2) * (-9.8 m/s^2) * T^2
where Dy is the vertical displacement (8.5 m), Viy is the initial vertical velocity that we need to find, T is the time (1.25 s), and -9.8 m/s^2 is the acceleration due to gravity.
Plugging in the values, we have:
8.5 m = Viy * 1.25 s + (1/2) * (-9.8 m/s^2) * (1.25 s)^2
Simplifying the equation, we get:
8.5 m = Viy * 1.25 s - 7.2265625 m
Vi = (8.5 m + 7.2265625 m) / 1.25 s
Vi = 13.7265625 m / 1.25 s
Vi = 10.98125 m/s
Therefore, the vertical component of the velocity (Vy) after 1.25 seconds is approximately 10.98125 m/s.
Now, we can find the resultant magnitude of the velocity (V) after 1.25 seconds using the Pythagorean theorem:
V = √(Vx^2 + Vy^2)
V = √(12 m/s)^2 + (10.98125 m/s)^2
V = √(144 m^2/s^2 + 120.5197266 m^2/s^2)
V = √(264.5197266 m^2/s^2)
V ≈ 16.26 m/s
Therefore, the velocity of the ball after 1.25 seconds is approximately 16.26 m/s.