A platform diver jumps vertically with a velocity of 4.2 m/s. The diver enters the water 2.5 s later. How high is the platform above the water?

41.13m

Explanation:

Given parameters:

Initial velocity  = 4.2m/s

Time taken = 2.5s

Unknown:

height of the platform = ?

Solution:

To solve this problem, we must use the appropriate motion equation.

   S = ut + gt²

u is the initial velocity

t is the time

g is the acceleration due to gravity

Now, input the parameters and solve;

   S = (4.2x2.5) + ( x 9.8 x 2.5²)

   S = 41.13m

a and b sometimes have a positive value

real, smaller and upside-down.   so the answer should be b.