Q&A for How to Find Initial Velocity

One of the difficulties people face is the unit for acceleration, which is meters per second squared. To understand this with a real-world example, think about a rock being dropped off a cliff. Acceleration due to Earth's gravity is 9.8 meters per second squared, so we can round up to 10 for this example. Neglecting friction, when you first let go of the rock, its velocity is zero. After one second of falling, the speed of the stone will be 10 meters a second. After two seconds of falling, the speed of the stone will be 20 meters a second. After three seconds of falling, the speed of this will be 30 meters per second. It gains 10 meters a second of speed for each second that it is falling.

Kinetic energy -> Kinetic energy 0.5mv^2 -> 0.5mv^2 0.5 x 12 x 2,5^2 = 0.5 x 0.06 x v^2 6 x 6.25 = 0.03 x v^2 37.5 = 0.03 x v^2 sqrt(37.5/0.03) = v sqrt(1250) = v v = 35.3 m/s

Initial velocity is 3.5. The equation is s = ut + 1/2at^2, where s - distance, u - inititial velocity, and a - acceleration.

You can't change the acceleration formula to one that gives you the initial velocity you want, as a=v/t. However, Vf=Vi+a.t is re-arranged. Vi=Vf-a.t, a=Vf-Vi/t, t=Vf-Vi/a.

Subtract the initial velocity from the final velocity, then divide the result by the time interval.

Assuming you are not including air resistance (which would make this problem far more difficult), the kinematic equations would be the usual s= (a/2)t^2+ vt+ d, where a is the acceleration vector, v is the initial velocity vector, and d is the initial position vector. Separating x (horizontal) and y (vertical) components and taking the initial speed to be "v" and the intial position to be d= (0, 0), we have x= v cos(30)t=(\sqrt{3}/2)v t and y= (-g/2)t^3+ v sin(30)= -4.9t^2+ (0.5)vt where v is the initial speed. Since the ball is to end up "20 meters away, the top edge is 5 meters above the throwing point", x= 20 and y= 5. Solve the two equations (\sqrt{3}/2)v t = 20 and -4.9t^2+

As soon as you stop, the final velocity is zero. Then, simply count backward to the start and divide by pieRx3.

The acceleration is how much the velocity of the car changes every second. If the velocity increased by 60 in 15 seconds, then in a second it would have increased by 4m/s.