Question

A stone is thrown vertically upword with a Velocity of 29.4 m/S. Calculate the first maximum height attained by the stone and time after which the stone will back to earth​

Answer 1

Answer:

Explanation:
Its given that u= 29.4m/s (approx 30m/s) so let's take 30 m/s as your initial velocity and a = 9.8m/s2(10m/s2 approx) which is acceleration due to gravity we don'tt know the time taken but we know the final velocity which will be 0 when it reaches the highest point or maximum height and its zero because acceleration  due to gravity retards the motion of stone traveling upwards
s=ut+1/2at2
this is the final formula we are going to use
to find t
we use v=u+at : 0=30+(-10)t.....(-10 because declarationn)
so -30/-10=t
t=3 seconds
now substitute 3 seconds in s=ut+1/at2
s= 30 * 3+1/2 *-10*(3)^2
s=90+ (-45)
s= 45 m
hence the maximum height attained will be 45 it doesnt matter before coming back on earth or up in the air when its velocity is 0 the height covered will be same, it applies for time too time taken while going up and coming down is equal