Question

what is the height attained by a stone thrown upward with velocity 50m/s​

Answer 1

Explanation:

Time to reach to the top point = (50–0)/9.8=5.10 sec. Ball will take exact same time to come back to ground level so total time required = 5.10x2=10.20 sec. That's the answer.

Answer 2

Answer:

${\huge{\red{\underline{\underline{Question}}}}}$

What is the height attained by a stone thrown upward with velocity 50m/s ?

${\huge{\red{\underline{\underline{Answer}}}}}$

${\pink{\underline{Given}}}$

Initial velocity(u) =50m/s

Final velocity(v)=0m/s as it reaches top its velocity is 0.

Acceleration(a)=-10m/s due to gravity directed downward

${\pink{\underline{To\ find:}}}$

• Height attend by stone=?????

${\pink{\underline{Solution}}}$

By applying third law of motion,

${v}^{2}={u}^{2}+2as$

Substituting the values;

${0}^{2}={(50)}^{2}+(2×-10×s)$

$=>0=2500+(-20)s$

$=>0-2500=-20s$

$=>\frac{-2500}{-20}=s$

$=>s=125m$

Hence,125 m is answer.