Question 65: Motion in 2D & Vectors

Question

A ball is projected from the ground with a speed 15 ms$$-$$1 at an angle $$	heta$$ with horizontal so that its range and maximum height are equal, then 'tan $$	heta$$' will be equal to :

Accepted Answer

To solve this problem, we will use the equations for the range and maximum height of a projectile. The range $$R$$ and maximum height $$H$$ of a projectile launched with speed $$u$$ at an angle $$	heta$$ can be expressed as follows:
Range:
$$R = \frac{u^2 \sin(2	heta)}{g}$$
Maximum height:
$$H = \frac{u^2 \sin^2(	heta)}{2g}$$
We are given that the range and maximum height are equal. So, we set these equations equal to each other:
$$\frac{u^2 \sin(2	heta)}{g} = \frac{u^2 \sin^2(	heta)}{2g}$$
We can cancel out the common terms $$u^2$$ and $$g$$ on both sides:
$$\sin(2	heta) = \frac{1}{2} \sin^2(	heta)$$
Using the double angle identity for sine, $$\sin(2	heta) = 2 \sin(	heta) \cos(	heta)$$, we substitute it into the equation:
$$2 \sin(	heta) \cos(	heta) = \frac{1}{2} \sin^2(	heta)$$
We can simplify this by dividing both sides by $$\sin(	heta)$$ (assuming $$	heta 
eq 0$$):
$$2 \cos(	heta) = \frac{1}{2} \sin(	heta)$$
Rearranging to get all terms on one side gives us:
$$4 \cos(	heta) = \sin(	heta)$$
Dividing both sides by $$\cos(	heta)$$, we get:
$$4 = 	an(	heta)$$
So, we find that:
$$	an(	heta) = 4$$
Thus, the correct answer is:
Option D: 4

Answer

$${1 \over 4}$$

Answer

$${1 \over 2}$$

Answer

2

Answer

4

Motion in 2D & Vectors

A ball is projected from the ground with a speed 15 ms$-1 at an angle \theta with horizontal so that its range and maximum height are equal, then 'tan \theta$' will be equal to :

Choose the correct answer: