Answer:
a) h(g) = 358,53 m
b) t = 8,16 s
c) t(t) = 16,71 s
Explanation:
Equations for vertical shooting are:
Vf = V₀ - g * t ; h = V₀*t - (1/2)*g*t² ; Vf² = V₀² - 2*g*h
And at maximum heigt Vf = 0 then
0 = V₀ - g * t
t = V₀/g V₀ = 80 m/s and g = 9,8 m/s²
t = 80 / 9,8 (s)
t = 8,16 s
Then 8,16 s is the time to get maximum height
If we plug t = 8,16 (s) in equation h = V₀*t - (1/2)*g*t²
we get: h (max) = (80)*8,16 - 0,5*9,8*(8,16)² (m)
h (max) = 652,8 - 326,27 m
h (max) = 326,53 m
Then relative to ground that height becomes
h(g) = 326,53 + 32
h(g) = 358,53 m
In order to get the time the arrow is in the air we proceed as follows:
a) for the arrow to be at the launched point will take the same time that from the launched point to the maximum height, and after that we have to find out the time the arrow takes from 32 m down to the ground level
Then
t(t) = 8,16 + 8,16 + tₓ (2)
Where tₓ is the time from 32 m height to ground
h = V₀*tₓ - (1/2)*g*tₓ² but since the arrow now is going down then we change the sign of the second term on the right side of the equation
32 = (80)*tₓ + 0,5 * 9,8 * tₓ² Note that when the arrow is at 32 m height the speed is again V₀ = 80 m/s
32 = 80*tₓ + 4,9*tₓ²
A second-degree equation for tₓ, solving it
4,9*tₓ² + 80*tₓ - 32 = 0
t₁,₂ = -80 ± √ 6400 + 627,2 / 9,8
t₁,₂ =( - 80 ± 83,8 ) / 9,8
there is not a negative time therefore we dismiss such solution and
t₁ = 3,8 / 9,8
t₁ = 0,39 s
And
t(t) = 8,16 + 8,16 + 0,39 s
t(t) = 16,71 s
