Question

NEED ASAP

A cannonball is fired at 85 m/s at 60° above the horizontal. For what period of time is the cannonball at height of 200 m or greater?

Answer 1

Time of flight, t = 2V₀sin/g
Here, V₀ is the initial velocity, sin is the component along Y-axis and g is the acceleration due to gravity.
V₀ = 85 m/s
sin= sin(60^o) = 0.866
g = 9.8 m/s²

t = (2 x 85 x 0.866) / 9.8
= 15.02 s
= 15 s

Time taken to reach 200 m is t₁. It can be calculated as follows.
s = V₀sint₁ + 1/2 at₁²
Here s = 200 m and
a = -g
Substituting values,
200 = 85 x 0.866 x t₁ + 1/2 x (-9.8) x t₁²
200 = 73.61 x t₁ - 4.9 x t₁²
4.9 x t₁² - 73.61 x t₁ + 200 = 0
Solving the quadratic equation, we will get
t₁ = [73.61 SQRT{ (-73.61)² - 4 x 4.9 x 200} ] / 2 x 4.9
= (73.61 38.71) / 9.8
= 11.46 or 3.56

Since time of flight is only 15 s, the time taken to reach 200 m cannot be 11 s. Therefore,
t₁ = 3.56 s

Period of time the cannonball at a height of 200 m or above = t - 2t₁
= 15.02 - (2 x 3.56)
=  7.89 s
= 7.9 s