Class 11 - Unit # 02 : Kinematics - Solved Numericals

Problem 1

A helicopter is ascending at the rate of 12 m/s. At a height of 80 m above the ground, a package is dropped. How long does the package take to reach the ground? (Ans: 5.43 s)

Given Data:

• Ascending velocity of helicopter (v_i) = 12 m/s
• Height of release point (h) = 80 m
• Acceleration due to gravity (g) = 9.8 m/s²
• Total time taken (t) = ?

Solution:

Note: Because the package is dropped from an ascending helicopter, it possesses an initial upward velocity of 12 m/s relative to the ground.

1. Upward Motion (From release point to highest peak):

At maximum height, the final velocity v_f = 0. Using the first equation of motion under gravity:

v_f = v_i - gt₁

0 = 12 - 9.8t₁ ⇒ 9.8t₁ = 12

t₁ =

12

9.8

= 1.22 s

Now, find the upward distance covered (S₁) during this phase:

2gS₁ = v_i² - v_f²

2(9.8)S₁ = (12)² - 0 ⇒ 19.6S₁ = 144

S₁ =

144

19.6

= 7.35 m

2. Downward Motion (From peak to the ground):

The maximum total height from the ground layout is:

Total Height (S₂) = h + S₁ = 80 + 7.35 = 87.35 m

During the fall from the peak, the initial velocity is 0. Using the second equation of motion:

S₂ =

1

2

gt₂²

87.35 =

1

2

× 9.8 × t₂² ⇒ 87.35 = 4.9 × t₂²

t₂² =

87.35

4.9

= 17.826

t₂ = √17.826 = 4.21 s

3. Total Time (t):

t = t₁ + t₂ = 1.22 s + 4.21 s = 5.43 s

Result:

The total time taken for the dropped package to reach the ground is 5.43 seconds.

Problem 2

Two tug boats are towing a ship. Each exerts a force of 6000 N, and the angle between the two ropes is 60°. Calculate the resultant force on the ship. (Ans: 10392.3 N)

Given Data:

• First Force (F₁) = 6000 N
• Second Force (F₂) = 6000 N
• Angle between forces (θ) = 60°
• Resultant Force (F_R) = ?

Solution:

According to the parallelogram law vector addition framework:

F_R = √F₁² + F₂² + 2F₁F₂ cos(θ)

Substituting values:

F_R = √(6000)² + (6000)² + 2(6000)(6000) cos(60°)

F_R = √3.6×10⁷ + 3.6×10⁷ + 7.2×10⁷(0.5)

F_R = √3.6×10⁷ + 3.6×10⁷ + 3.6×10⁷

F_R = √1.08 × 10⁸

F_R = 10392.3 N

Result:

The resultant tension vector force acting on the ship layout is 10392.3 N.

Problem 3

A car starts from rest and moves with a constant acceleration. During the 5th second of its motion, it covers a distance of 36 meters. Calculate: (a) acceleration of the car, and (b) the total distance covered by the car in 5 seconds. (Ans: 8 m/s², 100 m)

Given Data:

• Initial velocity (v_i) = 0 m/s
• Distance during 5th second (ΔS) = 36 m
• Acceleration (a) = ?
• Total distance in 5 seconds (S₅) = ?

Solution:

The distance covered in an interval is given by the difference between total distance at time n and time n-1.

(a) Finding Acceleration (a):

Using the distance equation S = v_it + ¹/₂at²:

Distance after 4 seconds (S₄):

S₄ = 0 +

1

2

a(4)² = 8a —— (Eq. i)

Distance after 5 seconds (S₅):

S₅ = 0 +

1

2

a(5)² = 12.5a —— (Eq. ii)

Distance covered precisely in the 5th second:

ΔS = S₅ - S₄

36 = 12.5a - 8a

36 = 4.5a ⇒ a =

36

4.5

a = 8 m/s²

(b) Finding Total Distance covered in 5 seconds (S₅):

Substitute the acceleration value back into Equation (ii):

S₅ = 12.5 × 8

S₅ = 100 m

Result:

The uniform acceleration of the car is 8 m/s² and the total cumulative distance traveled in 5 seconds equals 100 m.

Problem 4

Show proof that the horizontal range of a projectile at complementary angles is identical, using examples.

Theoretical Proof:

Complementary angles are pairs that sum up to 90°, represented as θ and (90° - θ). The horizontal range formula is:

R =

v_o² sin(2θ)

g

For the complementary angle layout (90° - θ):

R' =

v_o² sin[2(90° - θ)]

g

=

v_o² sin(180° - 2θ)

g

Since the trigonometric identity establishes that sin(180° - α) = sin(α):

R' =

v_o² sin(2θ)

g

= R

Verification Example (Using 30° and 60°):

Case 1: Angle θ₁ = 30°

R₁ =

v_o² sin(2 × 30°)

g

=

v_o² sin(60°)

g

=

v_o² (0.866)

g

Case 2: Angle θ₂ = 60°

R₂ =

v_o² sin(2 × 60°)

g

=

v_o² sin(120°)

g

=

v_o² (0.866)

g

Result:

Since R₁ equals R₂, it is proven that the horizontal range layouts at complementary launch angles are completely identical.

Problem 5

At what launch angle is the horizontal range of a projectile equal to its maximum height? (Ans: 76°)

Given Condition:

Horizontal Range (R) = Maximum Height (H)

Solution:

Recall the standard formulas for projectile motion matrix properties:

R =

v_o² (2 sinθ cosθ)

g

    and     H =

v_o² sin²θ

2g

Equating both equations as given by the problem statement condition:

v_o² (2 sinθ cosθ)

g

=

v_o² sin²θ

2g

Canceling matching terms (v_o², g, and one factor of sinθ) on both sides:

2 cosθ =

sinθ

2

Cross-multiplying and simplifying to form the tangent identity definition:

4 =

sinθ

cosθ

tanθ = 4

θ = tan^-1(4)

θ ≈ 75.96° ≈ 76°

Result:

The launch angle at which horizontal projectile range exactly equals maximum trajectory height is 76°.

Problem 6

A mortar shell is fired at a ground-level target 500 m away with an initial velocity of 90 m/s. What is the launch angle? [Ans: 18.6° or 71.4°]

Given Data:

• Horizontal Range (R) = 500 m
• Initial Velocity (v_o) = 90 m/s
• Acceleration due to gravity (g) = 9.8 m/s²
• Launch Angle (θ) = ?

Solution:

Using the horizontal range expression:

R =

v_o² sin(2θ)

g

Rearranging to isolate the sine function term:

sin(2θ) =

R × g

v_o²

Substituting values:

sin(2θ) =

500 × 9.8

(90)²

=

4900

8100

= 0.6049

Taking inverse sine value:

2θ = sin^-1(0.6049) ≈ 37.22°

θ₁ =

37.22°

2

= 18.61°

Since ranges match for complementary configuration pairs, the alternative launch angle option is:

θ₂ = 90° - 18.61° = 71.39°

Result:

The two possible launch configuration options to hit the target point are 18.6° (low trajectory) or 71.4° (high trajectory).