Question

In a projectile motion, the maximum
height reached equals the horizontal
range. The angle of the projectile with the
horizontal will be:
(A)
(B) tan (1)
tan
4.
(C) tan- (2) (D) tan(4)
-1​

Answer 1

Answer:

tan inverse (4)

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Answer 2

$\bold{Answer\;:-}$

• Given : The maximum height reached equals the horizontal range.

• To find : The angle of the projectile with the

horizontal range.

• Formulas :-

› Range, R = (u²sin2θ) / g

› Maximum height, H = (u²sin²θ) / 2g

• Solution :-

The question says that the maximum height reached equals the horizontal range. Thus,

› Range = Maximum height

› (u²sin2θ) / g = (u²sin²θ) / 2g

› (2u² × sinθ × cosθ) / g = (u²sin²θ) / 2g

› (2u² × sinθ × cosθ) / (u²sin²θ) = g / 2g

› (2 × cosθ) / sinθ = 1 / 2

› (2 × cosθ) × 2 = sinθ

› 4 × cosθ = sinθ

› sinθ / cosθ = 4

› tanθ = 4

› θ = tan^-1 (4)

• Extra :-

Range : The distance between the two points i.e. the initial point from where the projectile was thrown or launched to the point where it hit on the ground.

Maximum height : At this point, the forward velocity of the object is zero and as the name suggests, it the highest point in a trajectory.

$\bold{Hope\;it \; helps\;!}$