Question

An object at an angle such that the horizontal range is 4 times of the maximum height. What is the angle of projection of the object?

Answer 1

Final answer : Angle of Projection is 45°
STEPS:
1) According to the question,
Horizontal Range = 4 * (Maximum Height) ---(1)

2) We know that when prijectile is fired at angle ( a) ,then
H(max) = u^2( sin^2(a)) / 2g
Horizontal Range, R = u^2 sin(2a)/g
where u is initial velocity.

Using equation (1),
$\frac{ {u}^{2} \sin(2a) }{g} = 4 \times \frac{ {u}^{2} {sin}^{2} (a)}{2g} \\ = > \sin(2a) = 2 { \sin}^{2} (a) \\ = > 2 \sin(a) \cos(a) = 2 { \sin }^{2} (a) \\ = > \sin(a) = \cos(a) \\ = > a = 45 \degree \:$
Here, a is not equal to 0.
Hence, Projection angle is 45°.

Answer 2

Answer:i think question marked is not

correct

Answer of this question is

19°

Explanation:

We know that Range and height of projectile become equal at 76° so in this case if question asks that range should be 4 more then height

So answer will be like this

76 divide by 4

We get 19°