Question

A stone is thrown vertically upward from ground level with an initial velocity of 35.4 ? ? . A) At what subsequent times, relative to the starting point, is it 18 ? above ground level (i.e. on the way up and on the way down), and what are the corresponding velocities at these times? B) What is the highest point reached by the stone?

Answer 1

Initial Velocity of stone, u = 35.4 m/s

Part A

For this we will use 2nd equation of motion i.e.

where, h is the total displacement of the stone from its initial position to final position

u is the initial velocity with which stone is travelling initially

a is the acceleration of the stone (which is g (acceleration due to gravity) = 9.8 m/s² in our case as there is no other force acting on the stone than the gravitational force)

t is the time period of the motion

Now, we will be putting all quantities in the equation with +ve sign if they are measured or acting upward and -ve if measured or acting downward (Note :- Proper sign convention is very important in these equations)

Therefore putting, u = 35.4 m/s , h = 18 m , g = - 9.8m/s² (because gravity always acts downward)

This is an quadratic equation (equation having two solutions) of type At² + Bt + C

where A = 4.9 , B = - 35.4 and C = 18

and its two solutions will be given by below mentioned formula -

Two positive solutions of this equation means that the stone will pass through this particular point twice. Once on the way up and then on the way down.

Because t = 0.55 sec < t = 6.67 sec, Therefore First encounter of the stone with 18 m height will be at time t = 0.55 sec on the way up and then at time t = 6.67 sec on the way down, relative to the starting point respectively.

Now that we have time at which stone is at 18 m height, for finding the corresponding velocities at these points we will use 1st equation of motion i.e.

where, v is the final velocity of the stone

u is the initial velocity with which stone is travelling initially

a is the acceleration of the stone (which is g)

So putting all values with proper sign conventions i.e. u = 35.4 m/s , g = - 9.8 m/s² and t = 0.55 sec (1st case)

(Notice the sign of v, it is positive, that shows direction of final velocity of stone at t = 0.55 seconds is still upwards)

Now for t = 6.67 sec

(Notice the sign of v in this case now, it is negative, that shows direction of final velocity of stone at t = 6.67 seconds is downwards which makes sense as the stone is on the way down at this time)

Part B

Now for calculating the highest point reached by the stone we will use 3rd equation of motion

where, v is the final velocity of the stone

u is the initial velocity with which stone is travelling initially

a is the acceleration of the stone (which is g)

and, h is the total displacement of the stone from its initial position to final position

Now as the stone reaches the topmost point, It's final velocity (v) will be equal to ZERO at that particular instant. After that it will start coming down and height will decrease.

Therefore, for highest point, final velocity v = 0

Putting v = 0 m/s , u = 35.4 m/s , a = - 9.8 m/s²

Therefore highest point reached by the stone, h = 63.94 m