Question

please give me the answer​

Answer 1

In first Case,In Going Upward

Intial Velocity (u)=5 m/s

Final Velocity At Maximum Height (v)=0 m/s

Acceleration due to Gravity=-10 m/s

we take acceleration as Negative because when it was Going Upward .it was Working against gravity.

Use Equation of Motion

v²=u²+2as

Where v=Final Velocity

u=Intial Velocity

a=acceleration

s=Distance

so,

v²=u²+2as

0²=5²+2×-10×s

0=25-20s

25=20s

s=25/20

s=5/4

s=1.25 m

Distance=1.25m

We got

Intial Velocity=5 m/s

Final Velocity=0 m/s

Acceleration =-10 m/s

Use Equation of Motion

v=u+at

0=5+(-10)t

-5=-10t

-5/-10=t

0.5 sec=t

$\boxed{\mathbf{\huge{Height=1.25\:m}}}$

$\boxed{\mathbf{\huge{Time=0.5\:sec}}}$

Answer 2

Answer:-

Known Terms:-

Here we have,

u = Initial velocity

v = Final velocity

a = acceleration

s = Distance

t  = Time

 

Given:-

Initial velocity of stone, (u) = 5 m/s  

Final velocity of stone,  (v) = 0 m/s  

Deacceleration of stone,  (a) = – 10 m/s2  

To Find:-

Height attain by stone, (s) = ???

Time taken to reach the height, (t) = ???

Solution:-

1st Case

we know that v² - u²= 2as

Putting all values,

⇒ (0)²- (5)² = 2 × (-10) × s

⇒ $0 -25 = 20 s$

⇒ $20 s = 25$

⇒ $s \: = \frac{25}{20}$

⇒ $s \: = 1.25 \: m$

The height attained by the stone will be 1.25 m

2nd Case

As we know that:  v = u + at

Putting all values

⇒ $0 = 5 + (-10) t$

⇒ $0 = 5 - 10t$

⇒ $10t = 5$

⇒ $t = \frac{5}{10}$

⇒ $t = 0.5 s$

Time taken by the stone to reach at the top is 0.5 sec.