Question 181: Motion in 2D & Vectors

Question

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

Accepted Answer

Horizontal range
$$R = {{{u^2}\sin 2	heta } \over g}$$     ......(1)
Maximum height
$$H = {{{u^2}{{\sin }^2}	heta } \over {2g}}$$       .....(2)
According to the problem
R = H
$${{{u^2}\sin 2	heta } \over g}$$ = $${{{u^2}{{\sin }^2}	heta } \over {2g}}$$
$$ \Rightarrow 2\sin 	heta \cos 	heta  = {{{{\sin }^2}	heta } \over 2}$$
$$2\cos 	heta  = {{\sin 	heta } \over 2}$$
$$ \Rightarrow \cot 	heta  = {1 \over 4}$$
$$ \Rightarrow 	an 	heta  = 4$$
$$ \Rightarrow 	heta  = {	an ^{ - 1}}(4)$$

Answer

$$	heta $$ = tan$$-$$1$$\left( {{1 \over 4}} ight)$$

Answer

$$	heta $$ = tan$$-$$1(4)

Answer

$$	heta $$ = tan$$-$$1(2)

Answer

$$	heta $$ = 45o

Motion in 2D & Vectors

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

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