Question

a projectile has a range 80m and reaches a maximum height of 20m the angle of elevation at wich the projectile is fired​

Answer 1

Answer:

The angle of elevation is tan^-1 (1/2)

Explanation:

Let the angle of elevation be x,

R = u^2 sin2x / g

sin2x = Rg / u^2

sin2x = (80)(10) / u^2

sin2x = 800 / u^2

2sinxcosx = 800 / u^2

sinxcosx = 400 / u^2 ... (1)

H = u^2 (sinx)^2 / g

(sinx)^2 = Hg / u^2

(sinx)^2 = (20)(10) / u^2

(sinx)^2 = 200 / u^2 .... (2)

(2) / (1)

tanx = 1/2

x = tan^-1 (1/2)

Answer 2

Given:

A projectile has a range 80m and reaches a maximum height of 20m.

To find:

Angle of elevation ?

Calculation:

• Let Max height be H and range be R.

• After dividing the values of H and R, we get:

$\dfrac{R}{H} = \dfrac{80}{20}$

$\implies \dfrac{ \bigg( \dfrac{ {u}^{2} \sin(2 \theta) }{g} \bigg)}{\bigg( \dfrac{ {u}^{2} {\sin}^{2} ( \theta) }{2g} \bigg)} = 4$

$\implies \dfrac{ \sin(2 \theta) }{ {\sin}^{2} ( \theta) } = 4$

$\implies \dfrac{ 2\sin( \theta) \cos( \theta) }{ {\sin}^{2} ( \theta) } = 4$

$\implies \dfrac{ \cos( \theta) }{ \sin ( \theta) } = 2$

$\implies \cot( \theta) = 2$

$\implies \theta = { \cot}^{ - 1}( 2)$

So, angle of projection is cot^(-1)[2].