Question

A stone is thrown vertically upwards with a speed of 20m/s. how high will it go before it begins to fall? (take g= 10m/s^2)​

Answer 1

Answer:

20m

Explanation:

Lets take the kinematic equations v=u+at (1) and s=ut+(1/2)at^2 (2)

where u=initial velocity v= final velocity a= acceleration t=time taken. s= displacement travelled

Here we have been asked what the maximum height is.At the maximum point the stone comes to rest then starts falling.So at the highest point velocity is zero.

From question,

u=20 m/s v=0 m/s (at highest point) a= g = 10 m/s^2

from equation 1, v=u+at

0=20-gt (in upward throw acceleration is negative ) 20=10t

t=2

from equation 2, S= ut+(1/2)at^2

S=ut-(1/2)gt^2 ( upward throw a=negative)

S=20(2) -(1/2)(10) (2)^2

S= 40-5(4)

S=40-20

S=20 metres

Answer 2

Velocity at maximum height (v) = 0

Acceleration = -g

(-ve sign because it’s in the direction opposite to motion of stone)

$\implies \: {v}^{2} - {u}^{2} = 2as$

$\implies \: s = \frac{{v}^{2} - {u}^{2} }{2a}$

$\implies \: s = \frac{0 - {(20)}^{2} }{2 \times 10}$

$\implies \: s = \frac{ \cancel - \: 40 \cancel0}{ \cancel - \: 2 \cancel0} = 20 \: m$