Question

A ball rolls horizontally w/ a speed 6.89 mis off the edge of a platform. If the ball lands 7.84m from the point the ground directly below the edge of the platform. What is the height of the platform?

Answer 1

1) The horizontal compoenent of thevelocity does not change. The horizontal velocity is alys vx = 6.89 m/s. The horizontal distance is d = 7.84 m.The time taken to fall is

t = d/vx = 7.84/6.89 = 1.14 s

The height of the platform is

h = g t^2/2 = 9.8 x 1.14^2/2 = 6.34 m

The height is h = 6.34 m

2) Let the final temperature of the mixture be T

The heat gained by water is

H = mw cw (T-Tw)

mw = mass of water, cw = specific heat capacity of water, Tw = initial temperature of water

H = 3.3 x 4186(T - 33) =13813.8 (T-33 )

Similarly heat lost by oil is

H = 6 x 1970 (57-T) = 11820 (57 - T)

By conservation of energy, heat lost by oil = heat gained by water

11820 (57 - T) = 13813.8 (T - 33 )

673740 - 11820 T = 13813.8 T-455855.

or

(11820+13813.8) T = 673740+455855

or

25633.8 T. = 1129595

or T = 1129595/25633.8 = 44.07 C

The final temperature is T = 44.07 C

3) The difference in height is H = hA-hB = 12-7 = 5 m

The intial energy is Ei = m u^2/2

m is the mass and u = 8.8 m/s

The final energy at point B is Ef = mv^2/2 - mgH

Therefore by conservation of energy,

m u^2/2=mv^2/2 - mgH

or u = sqrt(v^2 + 2 g H) =sqrt(8.8^2 + 2 x 9.8 x 5) = 13.2 m/s

The final velocity is u = 13.2 m/s