Question

A ball is thrown with an initial upward velocity of 5 m/s.

Answer 1

Answer:

A ball is thrown at an initial height of 5 feet with an initial upward velocity at 29 ft/s. lets assume that  balls height h (in feet) after t seconds is give by:

h= 5 + 29t -16t^2

Explanation:

h= 5 + 29t -16t^2

 

a time when the ball's height will be 17 ft

 

17 = 5 + 29t -16t2

 

0 = -17 + 5 + 29t -16t2

 

0 = -12 + 29t - 16t2

 

Using the quadratic equation:

 

t = (-29±√(292-(4*(-16)*(-12))))÷2(-16)

 

 = (-29±√(841 - 768))÷(-32)

 

 = (-29±√(73))÷(-32)

 

 = (-29 + 8.544)÷(-32)   or   (-29 - 8.544)÷(-32)

 

 = (-20.456)÷(-32)         or   -37.544÷(-32)

 

 =  0.64                         or   1.17

 

So, the ball is at a height of 17 ft twice: once on the way up after 0.64 seconds and once on the way back down after 1.17 seconds.