Question

At what angle with horizontal should a ball thrown so that it's range R is related to the time of flight as R=5T²

Answer 1

Explanation:

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Answer 2

Answer:-

$\mathsf{\theta = 45^{\circ}}$

Given :-

Range is related to time of flight as $R = 5T^2$

To find :-

The angle of projection.

Solution:-

Since, ball is projected with certain angle.

The time taken by ball is given by :-

$\huge \boxed{T = \dfrac{2uSin\theta}{g}}$

The range covered by projected ball is given by :-

$\huge \boxed{R = \dfrac{u^2 Sin2\theta}{g}}$

And both are connected by relation,

R = 5T²

Put the values of R and T,

→$\mathsf{\dfrac{u^2Sin2 \theta}{g}=5 \left(\dfrac{2uSin\theta}{g}\right)^2}$

→$\mathsf{ \dfrac{u^2 Sin2 \theta }{g}= 5 \left(\dfrac{4u^2 Sin^2 \theta }{g^2}\right)}$

• Cancel out u² and g.

→$\mathsf{Sin 2 \theta = 5 \dfrac{4Sin^2 \theta}{g}}$

→$\mathsf{Sin2 \theta = \dfrac{5}{10}\times 4Sin^2 \theta }$

→$\mathsf{ \dfrac{2Sin\theta Cos\theta}{4Sin^2 \theta }= \dfrac{1}{2}}$

→$\mathsf{\dfrac{ Cos \theta}{2Sin\theta}= \dfrac{1}{2}}$

→$\mathsf{2 Cos \theta = 2Sin \theta}$

→$\dfrac{Sin\theta}{Cos \theta } = \dfrac{2}{2}$

$tan\theta = 1$

$tan\theta = tan45^{\circ}$

$\theta = 45^{\circ}$

hence,

The angle of projection is $\theta = 45^{\circ}$