A ball is thrown upwards from the ground with an initial . . .

Question : A ball is thrown upwards from the ground with an initial speed of u m/s. The ball is at the height of 80 m at two times. The time interval being 6sec between these two points. Find 'u'. Take g=10 m/s².

Doubt by Suhani

Solution : 

Here, the ball is thrown upward with initial velocity of u. After sometime the ball attains the height of 80 m from the ground then it reach the highest point, after the highest point, it starts coming down and again reaches the height of 80 m and then it comes back again at the ground. Hence, during it journey from bottom to top and then from top to bottom, the ball is at height of 80 m twice. The time interval between these two points is 6 sec. 

Now, Consider the motion of the ball from point B to C.

Here

Total Displacement (s) = 0

Time (T) = 6 sec
Let the velocity at point B = v m/s
Using Second Equation of motion
s=ut+½at²
0=v(6)+½(-10)(6)²
-6v=-5×36
v=5(36/6)
v=5×6
v=30 m/s

so, the velocity of the ball at point B is 30 m/s

[Note : You could also apply the formula of Time of flight which is T=2u/g if you remember it]

Now consider the motion of the ball from point A to B

Initial velocity at point A = u
Final velocity at point B = v = 30 m/s

Displacement (s) = 80 m

Using Third Equation of Motion

v²-u²=2as

(30)²-u²=2(-10)(80)
900-u²=-1600
-u²=-1600-900
-u²=-2500
u²=2500

u=√2500
u=±50 m/s
Rejecting the negative sign

u= 50 m/s

Hence, u = 50 m/s

Alternate Method : 

As we know that the ball is attaining the height (displacement) of 80 m twice so there must be two different times. Let these two times be t1 and t2.

Using Second Equation of Motion 

s=ut+½at²
80=ut+½(-10)t²
80=ut-5t²
5t²-ut+80=0

This is a Quadratic Equation. Hence, it has two values of time t.
Here
A = 5
B = -u
C = 80
D=B²-4AC
D=(-u)²-4(5)(80)
D=u²-1600

Using Quadratic Formula

There are two value of time t

t₁=t₂ = 6 sec (Given)

Hence, u = 50 m/s