Question

An object is thrown vertically upward with an initial velocity of 40m/s. Two second later another object is thrown upward with the same velocity.Find out followinga.At what height they meet?b. What is the time when they meet?c. What are the velocities of each object when they meet?​

Answer 1

hope this answer may correct

Answer 2

The answers are as follows-

• They meet at 75 meters from the ground

• They meet after 5 seconds

• The velocity first object is -10m/s

• The velocity second object is +10m/s

GIVEN

The initial velocity of the First and Second Object = 40 m/s

TO FIND

• Height when both objects meet.
• The time when they meet.
• Velocities of each object when they meet.

SOLUTION

We can simply solve the above problem as follows-

Let,

Time is taken when both objects meet = t seconds.

Time is taken by the Second object to meet the first object = t-2 seconds.

Applying the second equation of motion for the First and second objects respectively.

For object 1

Let the height covered by the First object be, S₁

$s = ut + \frac{1}{2} a {t}^{2}$

s₁= 40t + 1/2 + (-10 )×t×t. (acceleration due to gravity = 10 m/s²)

So,

s₁ = 40t - 5t². (Equation 1)

Let the final velocity of the first object be, V1

Applying the first equation of motion -

V1 = u + at

V1 = 40 - 10t. (Equation 2 )

For object 2

Let the distance covered by the object be, S₂

$s = ut + \frac{1}{2} a {t}^{2}$

$s = 40t + \frac{1}{2} - 10 \times (t - 2)^{2}$

Solving the above equation we get,

s₂= 40t - 5(t-2)². (equaEquation

Let the velocity at which the second object meets the first object be V₂

Applying the first equation of motion -

V₂ = 40 + (-10)(t-2) (equation 4)

We know that the height at which both the objects will meet is equal

Therefore,

s1 = s2

Putting the values of s1 and s2 in the above -

$40t - 5 {t}^{2} =40 (t - 2) + 5(t - 2)^{2}$

Solving the above equation-

$40t - 5t ^{2} = 40t - 80 + 5(t - 2)^{2}$

$80 = 5 {t}^{2} + 5(t - 2)^{2}$

16 = [t² -(t-2)²

16 = (t+t-2)(t-t-2)

16 = 2(2t-2)

2t-2 = 8

t = 10/2 = 5 sec

Hence, The time at which both the object will meet is 5 seconds.

Putting the value of t in equation (1)

s = 40×5 - 5×5²

= 200- 125

= 75 meters

Putting the value of t in equation 2 -

V₁ = 40 - 10×5

= 40-50 = -10

Hence, The velocity at which the second object meets the first object is 10 m/s in the downward direction.

Putting the value of t in equation 4

V₂ = 40 - 10(5-2)

= 40-30

= 10

The velocity of the second object is 10m/s upward direction.

Hence, The answers are as follows-

• They meet at 75 meters from the ground

• They meet after 5 seconds

• The velocity first object is -10m/s

• The velocity second object is +10m/s

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