Question

Derive the expression for the maximum height of a projectile

Answer 1

considering the figure,For maximum h at pont B, vertical component of velocity is 0. Maximum height is denoted by H=AB (from figure)
The Newton's second kinematical equation of motion is,
v$v^{2}$ =$u ^{2}$ + 2as ....(i)
For vertical motion of projectile,we have,
u=uy=usin$\alpha$      v=vy=0       , a=ay= -g , s=H 
Substituting in eaquation (i),we get,
0 = u$u^{2}$$sin^{2}$$\alpha$ - 2gH
.'. 2gH= $u^{2}$$sin^{2}$$\alpha$

.'.H=$u^{2}$$sin ^{2} \alpha /2g$

Answer 2

Answer:

H=(u sinθ)²/2g

Explanation:

v² = u² + 2as ----------------(1)

v = 0.

∴v^2 = 0 ----------------(2)

u = u sinθ ---------------(3)

a = -g  --------------------(4)

s = H ----------------------(5)

equating equations 2,3,4,5 in 1 we get;

0 = (u sinθ)² - 2gH

2gH = (u sinθ)²

∴H = (u sinθ)²/2g