A cricketer can throw a ball to a maximum horizontal distance of d. How high above the ground can the cricketer throw the same ball?

Question

A cricketer can throw a ball to a maximum horizontal distance of d. How high above the ground can the cricketer throw the same ball? (A) ( d /2) (B) d (C) 2 d (D) (5 d /2)

Tardigrade Admin · Accepted Answer

Let u is the velocity of projection, then R max =( u 2/ g )= d or u =√ gd Let h is the height upto which ball rise, then 0=u2-2 g h or h=(u2/2 g)=(g d/2 g)=(d/2)

Q. A cricketer can throw a ball to a maximum horizontal distance of $d$ . How high above the ground can the cricketer throw the same ball?

$\frac{d}{2}$

$d$

$2 d$

$\frac{5 d}{2}$

Let $u$ is the velocity of projection, then
$R_{m a x} = \frac{u ^{2}}{g} = d$ or $u = g d$
Let $h$ is the height upto which ball rise, then
$0 = u^{2} - 2 g h$
or $h = \frac{u ^{2}}{2 g} = \frac{g d}{2 g} = \frac{d}{2}$

Q. A cricketer can throw a ball to a maximum horizontal distance of d. How high above the ground can the cricketer throw the same ball?

2d​

d

2d

25d​

Let u is the velocity of projection, then Rmax​=gu2​=d or u=gd​ Let h is the height upto which ball rise, then 0=u2−2gh or h=2gu2​=2ggd​=2d​

Q. A cricketer can throw a ball to a maximum horizontal distance of $d$ . How high above the ground can the cricketer throw the same ball?

$\frac{d}{2}$

$d$

$2 d$

$\frac{5 d}{2}$

Let $u$ is the velocity of projection, then
$R_{m a x} = \frac{u ^{2}}{g} = d$ or $u = g d$
Let $h$ is the height upto which ball rise, then
$0 = u^{2} - 2 g h$
or $h = \frac{u ^{2}}{2 g} = \frac{g d}{2 g} = \frac{d}{2}$