A ball is thrown upward with an initial velocity of 14 m/s. Using the approximate value of g = 10 m/s2, how high above the ground is the ball at the following times?

(a) 0.90 s after it is thrown
(b) 2.10 s after it is thrown

h = -5 t^2 + 14 t ... meters

To find the height above the ground at a given time, we can use the kinematic equation:

h = v₀t + (1/2)gt²

where:
- h is the height above the ground
- v₀ is the initial velocity
- t is the time
- g is the acceleration due to gravity

Let's use this equation to find the answers for the given times.

(a) 0.90 s after it is thrown:
- Plug in the values:
h = (14 m/s)(0.90 s) + (1/2)(10 m/s²)(0.90 s)²
- Calculate:
h = 12.6 m + (1/2)(10 m/s²)(0.81 s²)
h = 12.6 m + (1/2)(8.1 m)
h = 12.6 m + 4.05 m
h = 16.65 m

So, the ball is approximately 16.65 m above the ground 0.90 seconds after it is thrown.

(b) 2.10 s after it is thrown:
- Plug in the values:
h = (14 m/s)(2.10 s) + (1/2)(10 m/s²)(2.10 s)²
- Calculate:
h = 29.4 m + (1/2)(10 m/s²)(4.41 s²)
h = 29.4 m + (1/2)(44.1 m)
h = 29.4 m + 22.05 m
h = 51.45 m

So, the ball is approximately 51.45 m above the ground 2.10 seconds after it is thrown.