Question

3. A ball is projected under gravity with velocity 20 m/s
at an angle with the horizontal such that horizontal
range is equal to the maximum height of projectile.
The range of projectile is (g = 10 ms?)
(1) 160 m
(2) 320 m
(3) 160÷17m
4)320÷17m​

Answer 1

Answer:

(320÷10) m

Explanation:

Data:

U = 20m/s; g = 10m/s²

To find the range

R = (U²Sin²∅)/g

And Maximum height

Hmax = (U²Sin²∅)/g

From the given question;

Hmax = Rx

U²Sin²∅/g = U²Sin2∅/g

⇒ Sin²∅ = Sin2∅

Where Sin2∅ = 2Sin∅Cos∅

⇒ Sin²∅ = 2Sin∅Cos∅

⇒ Sin∅ = 2Cos ∅

Divide both sides by Cos∅

⇒ Sin∅/Cos∅ = 2 {where Sin∅/Cos∅ = tan∅}

tan∅ = 2

∅ = arctan(2.000) = 63°

Therefore, the Range of the Projectile will be;

R = U²Sin2∅/g = 20²Sin126/10 = 400(0.8)/10 = (320÷ 10)m