Another Acceleration question?

Question

A child throws a rock upwards with a speed of 40.0 m/s. At the end of 10 seconds, (a) where is the stone and (b) at what speed is the stone moving?

My answers came out to:

(a) 890 m
(b) 138 m/s

The math seems right to me, but I feel icky about the high distance... Thanks.

Answer 1

Maria W got her signs mixed up (which I think is where you went wrong too). Her answers would be right if the child were throwing the rock _down_ at 40 m/s (say from the top of a building).

If you consider "up" to be the positive direction, then g is NEGATIVE, so change Maria's plus signs to minus signs:

v = v_initial + (-9.8m/s²)t
h = (v_initial)(t) + (-9.8m/s²)t²/2

(Alternatively you could have considered "down" to be negative; in that case g is positive but v_initial is negative. In either case, you need to subtract, not add).

v = 40 m/s – (9.8 m/s²)(10s) = -58 m/s
h = (40 m/s)(10s) – (9.8 m/s²)(10s)²/2 = -90 m

Notice that the final height is negative; that means the rock ends up 90 meters BELOW the child's location after 10 seconds. So, this can only happen if, say, the child is standing on the edge of a building or something. It only takes 8.2 seconds for the rock to go all the way up and then back down to the child's location.

Answer 2

The math is right, but the equations aren't the right ones to use. Energy equations are easier (kinetic and potential energy 1/2*m*v^2 and mgh respectively). You know when the ball leaves the childs hand it has all kinetic energy and when it reaches the top of the throw, stops and begins to fall it has all potential energy so you find those two values. The kinetic energy is found simply using 40 m/s as specified, after that it is constantly slowing down, until it falls back to the child and then it will have accelerated to the same speed. That KE value is the same as the rocks maximum PE value, which is at the top of the throw. I assumed the rock was about the same as a baseball, 0.142 kg. so then solve for h, which then, is the max height of the toss. This gives 81.6 m or about 244 ft. and the speed is again always changing but the max speed is 40 m/s or about 90 mph at the start of the throw and right before it lands again.

Answer 3

U r wrong ......beause u have not explored the other possibilities .actually in this case , the body will go to the top for some time and will be coming back at the instant when u need to calculate.
while going up , a = -9.8 take it as 10 for simplification .
v = u +at and u = 40
at top most point ...v =0
so , 0 = 40 - 10 *t
so , calculate it and by this u get time taken to reach to the top as 4 sec.
and , total height = 40 *4 - (1/2)10*(4^2) = 80 m

so for the next 6 sec remaining ..it will be dropping from the top .
while dropping from top..... u =0 , a = 9.8
so , height descended = ut + (1/2) a (t^2)
= 180 m .
this height is much greater then the height actually achieved.
So, the stone will be at the ground at the end of 10 seconds and will not be moving .

Answer 4

No, I think you're right...I guess the child is pretty strong?

s = ut + 1/2atsquared
s= 40(10) + !/2(9.8)(100) = 890 m

v = u + at
v = 40 + 10(9.8) = 138 m/s