free fall physics problem?

Question

I have been trying this problem for a while now and i can't seem to get the right answer. i know that part a is somewhere between 4 s. and 5s. and i got 45.1 for the speed.

A stone is thrown vertically upward with a speed of 13.0 m/s from the edge of a cliff 78.0 m high (Fig. 2-32).



(a) How much later does it reach the bottom of the cliff?
s
(b) What is its speed just before hitting?

Answer 1

It's a 2 part problem, you have to break it up into the trip up and the trip down.

For the first part (the trip up), use vi=13 m/s, vf=0 m/s and a=-9.8m/s^2 to solve for time and distance.

Next (the trip down), use vi=0 m/s, a=-9.8 m/s, and d=78 m (+ your distance from part 1) to solve for time on the way down.

Add the times from part 1 and 2 to find the answer for a. For b, use the variables from part 2 to solve vf. Watch your negatives!

Good luck!

Answer 2

Ok, you'd have to either know the kinematics formulas or know that they are derived from the integration of acceleration. The short way is to just know that, since the stone is only traveling in the y direction, all we are concerned with is the equation y=(1/2)at^2 + v(i)t + x(i), where t^2 is t squared, v(i) is the initial velocity, and x(i) is the initial distance. Thus when y=0 (when the stone hits the ground, it is 0 meters off the ground), we can solve for t. Since a=9.8 m/s^2, v(i)=13 m/s, x(i)=78 m, we can say 0= -4.9(m/s^2)t^2 +13 (m/s) t + 78 m.

Just do quadratic equation (-b +- radical( (b^2) - 4ac)) / 2a

t = -13 +- radical( 169 + 1528.8) / -9.8
t = (-13 - 41.204)/-9.8
t = 5.531 seconds (disregard negative answer).

The derivative....erm, the equation for velocity is v=at+v(i).
v = (-9.8)*(5.531) + 13
v = -41.204 m/s, so just before hitting, the velocity is 41.204 m/s
downward.

Hope this helps.

Answer 3

If the complete time from soar to max peak then right down to ground once more is two.eight s, the time to arrive max peak is a million/two of two.eight s on the grounds that of the symmetry of time, distance, and pace of "loose-fall" movement. d = a million/2gt² {in which d = max peak, g = nine.eighty one, t = a million.four} d = (zero.five)(nine.eighty one)(a million.four)² = nine.6 m ANS remark: method used is for the gap = d, as ball *falls* from relaxation to ground in a million.four s, and no longer the method for the ball because it *rises* from soar to max peak. The formulation are one-of-a-kind in each and every case, however the reply is the identical on the grounds that of movement symmetry.