Formual For Speed ?

Question

A cannon in a circus stunt fires a 750 N performer vertically to a height of 15.0m. What was the speed of the performer just as the cannon was fired ?

Answer 1

As you are given the performer's weight, I assume that you are meant to solve this using potential/kinetic energy conservation.

Potential energy at height h = Kinetic energy at beginning
U=mgh, mg = 750, h=15 -> U=11250 J
U=0.5 m v^2

v= sqrt(11250*2/(750/9.81) = 17.15 m/s

Although you could also do this using:
v^2-u^2 = 2as
where v=0 (stationary at the top) u is the initial speed, s is the height (15m) and a= -9.81 m/s^2 (acceleration due to gravity)

Again, this gives 17.15 m/s

Answer 2

Gravitational Potential Energy (GPE)
= mgh
= 750 *15
= 11250 J

K.E is max just as the cannon was fired, which was converted to GPE as the person gains altitude, which is max at the highest point. (assuming no air resistant)

So,
K.E= 0.5 (mv^2)
= 11250
v^2 = 30
v = sqrt 30

**The formula of speed is Change of Distance/ Time taken. But you do not need it here.

Answer 3

H = 1/2 a t^2

15 = 1/2 (9.8) t^2

t^2 = 15 * 2 / 9.8 T = 1.75 seconds

Vf = Vo + aT (Vf = 0 speed at top)
0 = Vo -9.8 * 1.75

Vo = 17.15 M/sec

Answer 4

Concurr with Steve.
Grant: this is correct, but we are given the weight, hence, the problem should be solved in terms of Energy.

Answer 5

speed=distance over time. you probably mean "velocity" which is mass*distance over time. with that your problem is missing the time factor in it. how fast did he get to the height of 15 m?

Answer 6

the formula for speed i learne is v=s/t
v= velocity
s= lenght
t=time

Answer 7

zero
Speed =distance/time
At that point, no time and no distance