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Suppose that a ball is thrown directly upward from a height of 2 meters with an initial velocity of 15 m/s. What is the formula for the height(t) of the ball t seconds after it is thrown?

 

At what time t will the ball in part C hit the ground?

 

at what time will the ball in part c reach its maximum height? what is this maximum height?

 Apr 12, 2021

Best Answer 

 #1
avatar+31586 
+2

y = 2  + 15 t   - 1/2 (9.81)t2  

 

 

y will equal 0 when the ball hits the ground

 

0 = 2 + 15t - 1/2 (9.81) t2          solve for t using Quadratic Formula    ( throw out negative value)

 

 

max height will occur at t = - b/2a =  - 15/ (2*4.905)    =  15/9.81 = ....... seconds

   use this value of 't' in the equation at the top (red)  to calculate the max height , y.....

 Apr 12, 2021
 #1
avatar+31586 
+2
Best Answer

y = 2  + 15 t   - 1/2 (9.81)t2  

 

 

y will equal 0 when the ball hits the ground

 

0 = 2 + 15t - 1/2 (9.81) t2          solve for t using Quadratic Formula    ( throw out negative value)

 

 

max height will occur at t = - b/2a =  - 15/ (2*4.905)    =  15/9.81 = ....... seconds

   use this value of 't' in the equation at the top (red)  to calculate the max height , y.....

ElectricPavlov Apr 12, 2021

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