A projectile is launched from ground level with speed u at an angle

of above the horizontal.
Find in terms of u,

o

and g by proving:

I. the maximum height H attained by the projectile,

II. the distance from the starting point at which the

projectile strikes the ground; this is called the range R

of the projectile.

III. The total time of flight

I. H = U² Sin²θ/2g

II. R = U² Sin 2θ/g

III. T = 2U Sinθ/g

Explanation:

We have a projectile motion with:

U = launch speed of projectile

θ = Angle of launch speed with horizontal surface

Ux = U Sinθ = Horizontal Component of Launch Speed

Uy = U Sinθ = Vertical Component of Launch Speed

H = Maximum Height attained by Projectile

R = Range of Projectile

t = Time to reach maximum height

T = 2t = Total Time of flight

I.

For H:

Considering the vertical motion of projectile, we use 3rd eqn. of motion:

2gh = Vf² - Vi²

here,

Vf = 0 m/s   (Because projectile stops at highest point)

Vi = Uy = U Sinθ

g = -g  (For upward motion)

h = H

Therefore,

-2gH = 0² - (U Sinθ)²

H = U² Sin²θ/2g

III.

For T:

Considering the vertical motion of projectile, we use 1st eqn. of motion:

Vf = Vi + gt

here,

Vf = 0 m/s   (Because projectile stops at highest point)

Vi = Uy = U Sinθ

g = -g  (For upward motion)

Therefore,

0 = U Sinθ - gt

t = U Sinθ/g

Therefore,

T = 2t

T = 2U Sinθ/g

II.

For R:

We neglect air resistance in our analysis, sue to which the horizontal velocity of projectile remains constant. Therefore:

S =Vt

where,

S = R

V = Ux = U Cosθ

t = T = 2U Sinθ/g

Therefore,

R = (U Cosθ)(2U Sinθ/g)

R = U²(2 SinθCosθ)/g

Since, Sin 2θ = 2 SinθCosθ

Therefore,

R = U² Sin 2θ/g

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer