Hi! I need help on my homework, I can't figure this out. subject: physics ?
A rock is thrown from a bridge 19.5 m above water with an initial speed of 17.5 m/s. Assuming negligible air resistance, what is the speed that the rock hits the water with, in meters per second?
Hint: "Since you don't know what angle the rock is thrown with, you don't know the individual components of velocity. This means using the conservation of energy will be a more productive approach."
- NCSLv 74 weeks agoFavourite answer
KE at impact = KE + PE at toss
½mv² = ½*m*(17.5m/s)² + m * 9.8m/s² * 19.5m
mass m cancels!
The rest solves to
v = 26.2 m/s
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- oubaasLv 74 weeks ago
Energy conservation shall apply :
Vf = √ Vi^2+2gh = √ 17.5^2+39*9.807 = 26.24 m/sec
- FiremanLv 74 weeks ago
BY the law of energy conservation:
=>KE(initial) + PE(initial) = KE(final)
=>1/2mu^2 + mgh = 1/2mv^2
=>u^2 + 2gh = v^2
=>v^2 = (17.5)^2 + 2 x 9.8 x 19.5
=>v = √[688.45]
=>v = 26.24 m/s
- Andrew SmithLv 74 weeks ago
if 1/2 m v^2 = 1/2 m u^2 + mgh ( conservation of energy ) then multiply both sides by 2/m ->v^2 = u^2 + 2gh ( which you really should remember) note that v^2 and u^2 become scalar not vector.
v= sqrt( u^2 + 2gh) = sqrt( 17.5^2 + 2 * 9.8 * 19.5) = 26.2 m/s
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- SpacemanLv 74 weeks ago
In order to use conservation of energy, you need to know the mass of the rock.
- 4 weeks ago
You know it is accelerating downwards at 9.8 m/s^2. If they don't define direction at all I would think you could come up with an answer if the rock was thrown directly down initial downward velocity of 17.5 m/s and another answer of a rock thrown horizontally with initial downward velocity of 0.