Question

a stone is thrown vertically upward reaches the maximum height in 3s. if the acceleration of the stonr be 9.8 m/s² calculate the initial velocity of the stone. Please solve this today is my exam.

Answer 1

Hey hi friend!!
Well this question is very easy
First let us see what is given to us
so we have,
The time taken to reach maximum height (t)=3sec
and acceleration due to gravity
TIP:- For smaller standards we usually have two known variables and we have to find the other unknown variable here we have given only one know variable and that is time we need one more to solve the question

Figuring out unknow variable:- usually  when we are given only one know variable we have to find another know variable from the question itself in this case we have found another know variable which is final velocity
Reason:- here the object is thrown upwards and the acceleration on it gravity which is constant hence we can apply kinematics for this equation. so when the object reaches maximum height its final velocity becomes 0 i.e. v-0
so thus we have found two variables t=3 sec and v=0 and a=g=9.8m/s^2 which is given

Thus by using the equation
v=u+at we get
0=u+(-9.8)3
0=u-29.4
u=29.4 m/s
hence the initial velocity of the object(stone) was 29.4 m/s

Hope this helped u :-)

Answer 2

t= 3 s
v= 0
a= -9.8
u= x

v= u + at
v= x + (-9.8) × 3
0 = x + (- 29.4)
so , u = 29.4m/s