Projectile from Height – Rankers Physics
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Projectile from Height

Question 1:

A particle is projected horizontally with a speed of 20/√3 m/s, from some height at t = 0. At what time will its velocity make 60° angle with the initial velocity

1 sec
2 sec
1.5 sec
2.5 sec
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Solution:

\[ tan \alpha = \frac{-gt}{u} \]

\[ tan (-60^{\circ })= \frac{-10t}{20/\sqrt{3}} \]

Solving t =2 sec

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Question 2:

Two tall buildings are 30 m apart. The speed with which a ball must be thrown horizontally from a window 150 m above the ground in one building so that it enters a window 27.5 m from the ground in the other building is :

\[2 ms^{-1}\]
\[6 ms^{-1}\]
\[4 ms^{-1}\]
\[8 ms^{-1}\]
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Solution:

Given:

- Horizontal distance between buildings \( x = 30 \, \text{m} \)
- Height difference \( h = 150 - 27.5 = 122.5 \, \text{m} \)
- Use \( h = \frac{1}{2} g t^2 \) to find time \( t \):

\[
122.5 = \frac{1}{2} \times 10 \times t^2 \quad \Rightarrow \quad t^2 = 24.5 \quad \Rightarrow \quad t = \sqrt{24.5} \approx 4.95 \, \text{seconds}
\]

Horizontal velocity \( v = \frac{x}{t} = \frac{30}{4.95} \approx 6.06 \, \text{m/s} \).

Thus, the required speed is \( {6.0 \, \text{m/s}} \).

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Question 3:

Two bullets are fired horizontally and simultaneously towards each other from roof tops of two buildings 100 m apart and of same height of 200 m, with the same velocity of 25 m/s. When and where will the two bullets collide? (g = 10 m/s²)

They will not collide
After 2 s at a height of 180 m
After 2 s at a height of 20 m
After 4 s at a height of 120 m
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Solution:

  • Horizontal motion: Bullets move towards each other with relative velocity 5050 m/s. Time to meet:

    t=10050=2 st = \frac{100}{50} = 2 \text{ s}

  • Vertical motion: Free fall equation:

    H=200−5t2=200−5(2)2=180 mH = 200 - 5t^2 = 200 - 5(2)^2 = 180 \text{ m}

Answer: (2) After 2 s at a height of 180 m.

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Question 4:

A ball is thrown horizontally from the top of a 80 m tall building with a speed of 20 m/s. Find the time it takes to reach the ground

4 sec
5 sec
6 sec
7 sec
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Solution:

Using formula for Time:

t=2H gt = \sqrt{\frac{2h}{g}} t=2×8010=16=4 st = \sqrt{\frac{2 \times 80}{10}} = \sqrt{16} = 4 \text{ s}

Correct Answer: (1) 4 s

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Question 5:

A ball is thrown horizontally with a speed of 20 m/s from a height of 80 m. Find the horizontal distance before hitting the ground.

60 m
70 m
80 m
90 m
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Solution:

Using the formula for range in horizontal projectile motion:

R=u ×2hgR = v_x \times \sqrt{\frac{2h}{g}} R=20×2×8010R = 20 \times \sqrt{\frac{2 \times 80}{10}} R=20×16=20×4=80 mR = 20 \times \sqrt{16} = 20 \times 4 = 80 \text{ m}

Correct Answer: (3) 80 m

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Question 6:

A helicopter flying at a height of 125 m and moving horizontally with a speed of 10 m/s drops a package. Find the horizontal distance it covers before hitting the ground.

20 m
30 m
40 m
50 m
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Solution:

Using the direct range formula for a projectile dropped from a height:

R=vx×2hgR = v_x \times \sqrt{\frac{2h}{g}}
R=10×2×12510R = 10 \times \sqrt{\frac{2 \times 125}{10}} R=10×25=10×5=50 mR = 10 \times \sqrt{25} = 10 \times 5 = 50 \text{ m}

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Question 7:

A helicopter is moving horizontally at a constant speed when it drops a package from a certain height. What will be the observed path of the package when viewed from the ground ?

A straight vertical line
A straight horizontal line
A parabola
A circle
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Solution:

\[ y = -\frac{g}{2 v_x^2} x^2 \]

This is of the form y=ax2y = ax^2, which represents a parabolic path. Hence, the object follows a parabolic trajectory when viewed from the ground.

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Question 8:

A helicopter is flying horizontally at a constant velocity. It releases a package, which falls freely under gravity. What will be the path of the package as seen by an observer inside the helicopter?

A Parabola
A Straight line
An Ellipse
A Circle
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Solution:

From the helicopter’s frame of reference, both the helicopter and the package have the same horizontal velocity. Since the package moves downward only due to gravity, it appears to fall straight down in a vertical line when observed from the helicopter.

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Question 9:

An object is projected horizontally from a 100 m high cliff with a speed of 10 m/s. What will be its velocity 1 second after projection?

10 m/s
20 m/s
10√2 m/s
10√3 m/s
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Solution:

Given:

vx=10v_x = 10

m/s,

g=10g = 10

m/s²,

t=1t = 1

s

Vertical velocity:

vy=gt=10v_y = g t = 10

m/s

Resultant velocity:

 

v=vx2+vy2=102+102=10√2  m/sv = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 10^2} = 14.1 \text{ m/s}

 

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Question 10:

An object is projected horizontally from a 45 m high cliff with a speed of 40 m/s. With what speed will it strike the ground? (Take g=10 m/s²)

20 m/s
30 m/s
40 m/s
50 m/s
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Solution:

  • Time to fall:
    t=2h/g=90/10=3t = \sqrt{2h/g} = \sqrt{90/10} = 3     

    s

  • Vertical velocity:
    vy=gt=10×3=30v_y = gt = 10 \times 3 = 30     

    m/s

  • Resultant velocity:
    v=vx2+vy2=402+302=50v = \sqrt{v_x^2 + v_y^2} = \sqrt{40^2 + 30^2} = 50     

    m/s

Answer: 50 m/s

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