If a person can throw a stone to maximum height of h metre vertically, then the maximum distance through which it can be thrown horizontally by the same person is

Question

If a person can throw a stone to maximum height of h metre vertically, then the maximum distance through which it can be thrown horizontally by the same person is (A) (h/2) (B) h (C) 2h (D) 3h

Tardigrade Admin · Accepted Answer

Using the equation of motion, we get h = (u2/2g) thus, u2 = 2gh So the maximum horizontal distance would be given by R textmax = (u2/g) ( where θ = 45°) Hence R textmax = 2h

Q. If a person can throw a stone to maximum height of $h$ metre vertically, then the maximum distance through which it can be thrown horizontally by the same person is

$\frac{h}{2}$

$h$

$2 h$

$3 h$

Using the equation of motion, we get
$h = \frac{u ^{2}}{2 g}$
thus, $u^{2} = 2 g h$
So the maximum horizontal distance would be given by
$R_{max} = \frac{u ^{2}}{g}$ ( where $θ = 4 5^{\circ})$
Hence $R_{max} = 2 h$

Q. If a person can throw a stone to maximum height of h metre vertically, then the maximum distance through which it can be thrown horizontally by the same person is

2h​

h

2h

3h

Using the equation of motion, we get h=2gu2​ thus, u2=2gh So the maximum horizontal distance would be given by Rmax​=gu2​( where θ=45∘) Hence Rmax​=2h

Q. If a person can throw a stone to maximum height of $h$ metre vertically, then the maximum distance through which it can be thrown horizontally by the same person is

$\frac{h}{2}$

$h$

$2 h$

$3 h$

Using the equation of motion, we get
$h = \frac{u ^{2}}{2 g}$
thus, $u^{2} = 2 g h$
So the maximum horizontal distance would be given by
$R_{max} = \frac{u ^{2}}{g}$ ( where $θ = 4 5^{\circ})$
Hence $R_{max} = 2 h$