Question

A baseball is thrown in parabolic arc its position above the ground at a given point in time can be represented by a quadratic function p(t)= 1/2gt^2 + v0t +p0 where t > 0, g is -32ft/sec/sec, V0 is additionally velocity and P0 SH initial position above the ground. If the ball is thrown straight up 16ft/sec when it was 5ft above the ground how high did it go

Answer 1

Answer:

13 ft.

Step-by-step explanation:

Since  p(t)= 1/2gt² + v₀t +p₀ and it us given that v₀ = 16 ft/s, p₀ = 5 ft and g = -32 ft/s².

Since the ball is thrown straight up, it does not follow a parabolic path.

So, If the ball is thrown straight up 16ft/sec when it was 5ft above the ground, then its height above the ground h(t) = v₀t +p₀. Substituting v₀ = 16 ft/s, p₀ = 5 ft , we have

h(t) = v₀t +p₀

Using v = v₀ + at, we calculate the time it takes to reach maximum height. At maximum height, v = 0, so,

v = v₀ + at

v - v₀ = at  

t = (v - v₀)/a where  a = g

So, t = (0 - 16 ft/s)/-32 ft/s²

=  - 16 ft/s/-32 ft/s²

= 1/2 s

= 0.5 s

Substituting t = 0.5 s, v₀ = 16 ft/s and p₀ = 5 ft into h(t), we have

h(t) = v₀t +p₀

h(t) = 16 ft/s × 0.5 s + 5 ft

h(t) = 8 ft + 5 ft

h(t )= 13 ft

The ball will go 13 ft above the ground.