Question

A projectile is launched with an initial velocity of (40 m/s), at an angle of (30°) above

horizontal. What is the maximum height (ymax) reached by the projectile?
(Use g=10 m/s2).

Answer 1

330.5 m

Step-by-step explanation:

In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .

The maximum height will be calculated as;

`h=(v^2_isin^2\alpha )/(2g)`

where ∝ is the angle of launch = 30°

vi= initial launch velocity = 40 m/s

g= 10 m/s²

h= 40²*sin²40° / 2*10

h={1600*0.4132 }/ 20

h= 661.1/2 = 330.5 m