Anonymous
Anonymous asked in Science & MathematicsPhysics · 4 weeks ago

# 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."

Relevance

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

• Energy conservation shall apply :

Vf = √ Vi^2+2gh = √ 17.5^2+39*9.807 = 26.24 m/sec

• 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

• 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