Question

You toss a rock of mass m vertically upward. Air resistance can be neglected. The rock reaches a maximum height h above your hand. What is the speed of the rock when it is at height (a) h/4 and (b) 3h/4

Answer 1

Answer:

Explanation:

mass of rock = m

height reached = h

let the rock is thrown with velocity u .

At maximum height the velocity of rock is zero.

By using third equation of motion

v² = u² - 2gh

0 = u² - 2 gh

h = u² / 2g    .... (1)

(a)

Let the velocity is v at height h/4.

Again using third equation of motion

v² = u² - 2g h/4

v² = u² - 2 g x u²/8 g = 3u²/4

v = 0.866 u

Substitute the value of u from equation (1)

v = 0.866 x √2gh = 3.84 √h

(b) Let v be the velocity at height 3h/4

Again using third equation of motion

v² = u² - 2gx 3h/4

v² = u² - 6 g x u²/8 g = u²/4

v = 0.5 u

Substitute the value of u from equation (1)

v = 0.5 x √2gh = 2.2 √h

Answer 2

Answer with Explanation:

We are given that

Mass of rock=m

Maximum height=h

a.At maximum height, velocity,v=0

We know that

$v^2=u^2-2gh$

$0+2gh=u^2$

$u^2=2gh$

Height,h=h/4

Again,$v'^2=u^2-2g\times \frac{h}{4}$

$v'^2=2gh-\frac{gh}{2}=\frac{4gh-gh}{2}=\frac{3gh}{2}$

$v'=\sqrt{\frac{3gh}{2}}=\sqrt{\frac{3\times 9.8 h}{2}}=3.83\sqrt h$

Where $g=9.8 m/s^2$

b.When height,h=3h/4

$v'^2=u^2-2gh$

$v'^2=2gh-2g\times \frac{3h}{4}=2gh-\frac{3gh}{2}=\frac{4gh-3gh}{2}=\frac{gh}{2}$

$v'=\sqrt{\frac{9.8h}{2}}=2.2\sqrt h$

$v'=2.2\sqrt h$