A ball is dropped from a ballon going up at a speed of 7m/s . If the ballon was at 60m at time of dropping the ball how long the ball take to reach ground level

Question

A ball is dropped from a ballon going up at a speed of 7m/s . If the ballon was at 60m at time of dropping the ball how long the ball take to reach ground level .

Answer 1

Hi Tony,

 

To answer this question, you must first know the equation of motion:

 

x(t)=(1/2)a₀t^2+v₀t+x₀

 

where a₀ is the initial acceleration, v₀ is the initial velocity, and x₀ is the initial position.

 

Now, a₀ is the acceleration due to gravity, so a₀=-9.8 m/s^2. The initial velocity is given as v₀=7 m/s. The intial position is x₀=60 m (the position of the ball x(t) when t=0). So, plugging these values into our equation of motion x(t) we have:

 

x(t) = (1/2)*(-9.8)*t^2 + 7*t + 60

 

or

 

x(t) = -4.9t^2 + 7t + 60

 

To find how long until the ball reaches ground level, we simply solve x(t)=0. Using the quatratic formula, we have:

 

t=(-7 +/- sqrt( 7^2 - 4*(-4.9)*60 ) )/( 2*(-4.9) )

 

Taking this in pieces, we have two answers:

 

t=(-7+sqrt(49+1176))/-9.8

 =-2.86 s

 

and

 

t=(-7-sqrt(49+1176))/-9.8

 =4.29 s

 

Since we only care about what happens after the ball has left the balloon, we take the positive value        t=4.29 s. The ball will be at ground level after 4.29 seconds.