## How would i derive the equation?

Question 1:
A ball is dropped from the top of a building and it strikes the ground with a speed of v m/s. From ground level, a second ball is thrown straight upward with the same initial speed of v m/s at the same instant that the first ball is dropped. Ignoring air resistance, determine at what height above the ground the two balls will pass.

Question 2:
A cab driver picks up a customer and delivers her 2.0km away after driving along a striaght route. In making the trip, the driver accelerates uniformly to the speed limit and upon reaching it, decelerates uniformly immediately. The magnitude of the deceleration is three times the magnitude of the acceleration. Calculate the distance travelled during the acceleration and deceleration phases of the trip.

Question 3:
A vehicle spped trap is set up with pressure activiated strips across the highway, 110 meters apart. a car is diring along at 33m/s in a area where the speed limit is 21m/s. At the instant the vehicle activates the first strip, the driver begins slowing down uniformly. What is the minimum deceleration need in order that the average speed of the vehicle between the strips does not exceed the speed limit.

I have no idea on how to start on these problems... so i have tried it, but i can't get a clear idea to begin with.

Thanks.
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 Recognitions: Homework Help Science Advisor For #1 you can express the height of the building in terms of the final velocity v. You can then write two equations for the vertical positions of the balls in terms of their initial positions relative to the ground, initial velocities, acceleration, and time. Find the time and the vertical height when they have the same vertical height.
 umm well i got two equations, the first one would be D=1/2at^2, (downward ball after it has reached the highest point then free fall.), the second would be D=v/mst-1/2at^2, upward motion slows down by gravity. then?

Recognitions:
Homework Help

## How would i derive the equation?

 Quote by Tensaiga umm well i got two equations, the first one would be D=1/2at^2, (downward ball after it has reached the highest point then free fall.), the second would be D=v/mst-1/2at^2, upward motion slows down by gravity. then?
Express both of your equations in the same form and with the same reference point. Use the general equation

y = y_o + v_o*t - ½*g*t²

For the ball that is dropped, you need to find y_o. You can do that knowing that when it hits the ground (y = 0) its velocity is -v, and it started from rest, so the change in velocity is -v - 0 and that change is due to the acceleration acting for some time -gt. An equatioin has already been worked out based on this reasoning that involves the velocities, acceleration, and distance moved. You can use that equation if you prefer. Once youu get the two equation in the form written above. All you have to do is find the t and y value that makes the y and t in both equations the same.