the energy of the ball when it leaves the spring is ... m * g * h
the ball's velocity is ... 1/2 m v^2 = m g h
... V = √(2 g h)
at a 33º angle
... horizontal v = V cos(33º)
... vertical v = V sin(33º)
use the vertical component to find the flight time (t)
... t = 2[V sin(33º) / g]
use the horizontal component to find the range (r)
... r = V cos(33º) * t
A small steel ball bearing with a mass of 27 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.43 m. Calculate the horizontal distance the ball would travel if the same spring were aimed 33 o from the horizontal.
I'm doing something wrong in my calculations, is it possible for you to help out with detailed calculations? Thank you!
2 answers
m = 0.027 kg , S = speed
but does not matter, call it m
m g h = (1/2) m S^2
9.81 * 1.43 * 2 = S^2
S = 5.3 m/s straight up
give it the same speed at 33 deg above horizon
Vi = 5.3 sin 33 = 2.89 m/s up component
Vertical problem:
v = Vi - 9.81 t
top when v = 0
t = 2.88/9.81 = 0.294 seconds upward
another 0.294 to fall
so 0.588 seconds in air
Horizontal problem
speed = u = 5.3 cos 33 = 4.19 m/s
for 0.588 s
so
2.47 meters
but does not matter, call it m
m g h = (1/2) m S^2
9.81 * 1.43 * 2 = S^2
S = 5.3 m/s straight up
give it the same speed at 33 deg above horizon
Vi = 5.3 sin 33 = 2.89 m/s up component
Vertical problem:
v = Vi - 9.81 t
top when v = 0
t = 2.88/9.81 = 0.294 seconds upward
another 0.294 to fall
so 0.588 seconds in air
Horizontal problem
speed = u = 5.3 cos 33 = 4.19 m/s
for 0.588 s
so
2.47 meters
1 answer