Question

A bomb is thrown at a angle of 30 with 10 m/s with respect to ground. What will be range for

60 with respect to ground? Use (g=10) ​

Answer 1

Given :-

▪ A bomb is thrown at an angle of, θ = 30° with a initial velocity of u = 10 m/s with respect to the ground.

To Find :-

▪ Range of the projectile with respect to the ground.

Solution :-

The given motion of the bomb is a projectile motion, so the formula of range can be applied here. We have,

• θ = 30°
• g = 10 m/s²
• u = 10 m/s²

In the Question, It is written with respect to ground. Whenever we are given these kind of statements, just the value given instead of getting into the relative velocity topic.

So, Range of the Bomb can be calculated as,

⇒ R = u² sin 2θ / g

⇒ R = (10)² sin (2×30°) / 10

⇒ R = 10 sin 60°

⇒ R = 10 × √3 / 2

⇒ R = 5 × √3

⇒ R = 5 × 1.73 [ take, √3 = 1.73 ]

⇒ R = 8.65 m

Hence, The range of the bomb is 8.65 m.

Answer 2

Correct question:

A bomb is thrown at an angle of 30° with 10 m/s with respect to ground. What will be range for

it with respect to ground? Use (g=10)

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Given:

• Angle of projection = $\theta$ = 30°
• Initial velocity = u = 10m/s
• g = 10m/s2

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Need to find:

• Range =?

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Solution:

When an object is thrown making an angle with the ground then it goes in a parabolic path and that motion is called projectile motion.

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We know, For projectile motion :

$R= \dfrac{ {u}^{2} sin \: 2\theta }{g}m$

$\implies \: R = \dfrac{ {10}^{2} \times \sin(2 \times 30) }{10}m$

$\implies \: R = \dfrac{100 \times \sin(60) }{10} m$

$\implies \: R = 10 \times \dfrac{ \sqrt{3} }{2} m$

$\red{\bold{\large{\boxed{ \implies \: R = 5 \sqrt{3} m}}}}$

Now putting value of root 3= 1.732 we get,

⟹ R = 5 × 1.732 m

$\red{\bold{\large{\boxed{ \implies \: R = 8.66 m}}}}$