Projectile from Height - NEET Physics Questions
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Projectile from Height

Question 1:

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 :

1. \[2 ms^{-1}\]
2. \[6 ms^{-1}\]
3. \[4 ms^{-1}\]
4. \[8 ms^{-1}\]
View Answer

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 \( {6.0 \, \text{m/s}} \).

Question 2:

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?

1. 10 m/s
2. 20 m/s
3. 10āˆš2 m/s
4. 10āˆš3 m/s
View Answer

Given:

vx=10v_x = 10

m/s,

g=10g = 10

m/sĀ²,

t=1t = 1

s

Vertical velocity:

vy=gt=10v_y = g t = 10

m/s

Resultant velocity:

 

v=vx2+vy2=102+102=10āˆš2 Ā m/sv = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 10^2} = 14.1 \text{ m/s}

 

Question 3:

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/sĀ²)

1. 20 m/s
2. 30 m/s
3. 40 m/s
4. 50 m/s
View Answer
  • Time to fall:
    t=2h/g=90/10=3t = \sqrt{2h/g} = \sqrt{90/10} = 3     

    s

  • Vertical velocity:
    vy=gt=10Ɨ3=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