Question

A ball is dropped from a building having a height 50m. Calculate the time it takes to reach the ground.

Answer 1

S = ut + 1/2 * g * t^2 ( where S is distance and t is time)

50 = 0 + 1/2 * 9.8 * t^2

:. t = 3.19 s

Answer 2

Any object will accellerate due to gravity -9.81 m/s^2. By integrating this acceleration over time, one obtains the velocity at any time t.

v = gt + v0. In this case, the ball started at rest, so the initial velocity v0 is zero.

By integrating velocity over time, one obtains the position y at any time t.

y = 1/2gt^2 + y0. We know that y0, the initial height of the ball, is 50 m.

So, when does the ball hit the ground (y = 0). Solve for t:

0 = 1/2 (-9.81 m/s^2) t^2 + 50 --> 100s^2 = 9.81 t^2 --> t = 3.2 s

You can answer the second question by examining the energy of the system. The kinetic energy when the ball hits the ground is determined by it's velocity. 1/2 mv^2. The velocity at t = 3.2s is v = g*3.2 = 10.2 m/s.

Uk = 52*m J/kg

The problem states that the ball looses half it's energy to the impact. So the new kinetic energy is 26*m J/kg. How high will it go? All of the kinetic energy will go to potential energy according to the equation Up = mgh. So,

Uk = Up --> 52m = mgh --> h = 52/g = 5.3 meters.

Answer 3

let time be t.

then 0.5*9.8*t^2=50

t=3.194 sec

Answer 4

s = ut+(1/2)*g*(t^2)

50 = 0 + (1/2)*9.81*(t^2)

t = sqrt(50*2/9.81) = 3.1927 seconds

Answer 5

s=ut+1/2 gt^2 here S is height which is given as 50

and initial velocity is 0 so ut=0 , therefore, s=1/2 *gt^2

50=0+ 1/2*9.8t^2

t^2= 50/4.9

t=3.194sec

Answer 6

s=ut + a(t^2)/2

here u=0,a=9.8 or 10(approax)

and s=50

after calculating .. t=3.2 s (approax)