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

Question

A ball is projected from the ground with a speed 15 ms -1 at an angle θ with horizontal so that its range and maximum height are equal, then'tan θ ' will be equal to (A) (1/4) (B) (1/2) (C) 2 (D) 4

Tardigrade Admin · Accepted Answer

R = H (2 v x × v y / g )=( v y 2/2 g ) v x =( v y /4) ; u cos θ=( u sin θ/4) tan θ=4

Q. A ball is projected from the ground with a speed $15 m s^{- 1}$ at an angle $θ$ with horizontal so that its range and maximum height are equal, then'tan $θ$ ' will be equal to

$\frac{1}{4}$

$\frac{1}{2}$

2

4

$R = H$
$\frac{2 v _{x} \times v _{y}}{g} = \frac{v _{y}^{2}}{2 g}$
$v_{x} = \frac{v _{y}}{4}; u cos θ = \frac{u s i n θ}{4}$
$tan θ = 4$

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

41​

21​

2

4

R=H g2vx​×vy​​=2gvy2​​ vx​=4vy​​;ucosθ=4usinθ​ tanθ=4

Q. A ball is projected from the ground with a speed $15 m s^{- 1}$ at an angle $θ$ with horizontal so that its range and maximum height are equal, then'tan $θ$ ' will be equal to

$\frac{1}{4}$

$\frac{1}{2}$

$R = H$
$\frac{2 v _{x} \times v _{y}}{g} = \frac{v _{y}^{2}}{2 g}$
$v_{x} = \frac{v _{y}}{4}; u cos θ = \frac{u s i n θ}{4}$
$tan θ = 4$