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Physics help on motion in 1-d please help

  1. Oct 2, 2003 #1
    A stone is thrown vertically upward with a speed of 10 m/s from the edge of the cliff 65 m high. It reaches the ground sometime later

    a) what is the acceleration and velocity at the highest point. (I got acceleration=9.8 m/s^2 and velocity=o)

    b)What is the displacement for the freefall portion of the trip? (I got 0 m)

    c) what is the final free fall velocity?

    d) how long is the stone in freefall?

    2)A car starts from rest and accelerates uniformily to a speed of 30 m/s in 10s. The driver maintains this speed for another 10s.

    What is the displacement for the entire 20s trip?

    Thanx a lot
  2. jcsd
  3. Oct 2, 2003 #2
    Basically, as the stone is thrown upwards, it is decelerating due to the acceleration due to gravity (g). Eventually, the velocity will reach zero at which time it begins the decent back to earth, again due to gravity. Therefore:

    a) You are correct :smile: .Acceleration at the highest point = -9.8 meters per second per second and the velocity is 0.

    b)We will assume that the stone falls past the starting point to the bottom of the cliff.

    Here we need to determine how far the stone goes up using the kinematic equation: velocity final squared = velocity initial squared + 2ay. We are using y as the variable as the motion is vertical rather than horizontal. Solving for y = (Vfinal^2 - Vinitial^2) / 2g. I calculate about +5.1 meters from the top of the cliff. +5.1 meters + 65 meters (the initial starting heighth) = 70.1 meters (the total length of the free fall to the ground).

    c) Now that y has been determined, a second equation of y= (gt^2)/2 will give you the time. Note: the actual equation is y = Vinitial * t + (gt^2)/2, but Vinitial is = to zero. Solving for t will yield the time for the free fall. Vfinal = Vinitial + at will give you your final velocity.

    d) See above for time in freefall.

    2) For problem two we need to divide the trip it into its components. The first 10 sec will be 1 and the second 10 seconds will be 2.

    x1 = average velocity1 * t1 = (Vfinal1 - Vinitial1)t1/2. Plugging in the values results in (30m/s - 0m/s)*10s/2 = 150 meters.


    For this segment the velocity remains constant; therefore we can just multiply the velocity by the time to get the total distance for the second segment (300 m).

    x1 + x2 = total displacement = 450m
    Last edited: Oct 2, 2003
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