A 3.00 kg ball is dropped into a lake at the top of the 200m deep lake . The water exerts a upwards buoyant force on the ball the ball has a velocity of 12.6m/s (down) just before it hits the bottom of the lake. Calculate the acceleration.

Accepted Solution

The acceleration may be determined by the formula $$v^2=2as$$ where v is the final speed, a is the acceleration and s is the distance (or depth of the water). Hence $$v=12.6m/s,\quad s=200m,\quad a=?$$ Substitute into the equation $$v^2=2as$$ $$12.6^2=2a(200)$$ $$158.76=400a$$ $$a=\frac{158.76}{400}\approx 0.40$$ Hence the acceleration is about $$0.4m/s^2$$