You plan to throw stones by using a sling of length 1.8 m which you whirl over your head. Suppose you wish to throw a stone a distance of 29 m. What must be the centripetal acceleration of the stone just before its release if it is to reach this distance? Assume that the release height is 2.8 m.
So I assume the angle of whirling is parallel to the ground.
Time to fall 2.8 m: 2.8=4.9t^2
solve for time in air.
29m=Vi*timeinair
solve for Vi.
then, centripetal acceleration
thanks that makes sense
To find the required centripetal acceleration of the stone, we can use the formula:
centripetal acceleration (a) = (v^2) / r
where:
- v is the final velocity of the stone just before release
- r is the radius of the circular path of the stone (in this case, the length of the sling, 1.8 m)
To find v, we will use the equation of motion:
v^2 = u^2 + 2as
where:
- u is the initial velocity (which is 0 since the stone is being released from rest)
- a is the acceleration (in this case, the centripetal acceleration)
- s is the displacement (in this case, the distance the stone needs to travel, 29 m)
We need to find a in order to find v, but to use the equation of motion, we also need to calculate the time it takes for the stone to travel the distance of 29 m.
To find the time (t), we can use the equation:
s = ut + (1/2)at^2
where:
- s is the displacement (29 m)
- u is the initial velocity (0 m/s)
- a is the acceleration (unknown)
- t is the time (unknown)
The equation becomes:
29 = (1/2)at^2
Now, let's solve for t:
58 = at^2
t^2 = 58/a
t = sqrt(58/a)
Plugging this value of t into the equation of motion:
29 = 0 + (1/2)a(sqrt(58/a))^2
29 = 0 + (1/2)a(58/a)
29 = 29/2
This equation tells us that a = 2 m/s^2, which gives us the required centripetal acceleration for the stone just before its release in order to reach a distance of 29 m.
Note: The release height of 2.8 m is not relevant to finding the centripetal acceleration. It only affects the vertical motion of the stone.