Question

A body is projected with initial velocity of 40 m/s and at an angle of 30°with the horizontal. Calculate (i) time of flight and (i
i) horizontal range of projectile.(assume g= 10 m/s^2)​

Answer 1

Answer:-

→ Time of flight is 4 seconds .

→ Horizontal range is 138.56 m .

Explanation:-

We have :-

→ Initial velocity of the body = 40 m/s

→ Angle of projection = 30°

→ Acceleration due to gravity = 10 m/s²

________________________________

• sin 30° = 0.5

• sin 60° = √3/2

We know that :-

T = 2usinθ/g

Substituting values, we get :-

⇒ T = 2 × 40 × sin30°/10

⇒ T = 80 × 0.5/10

⇒ T = 40/10

⇒ T = 4 seconds

Now :-

R = u²sin2θ/g

Substituting values, we get :-

⇒ R = (40)²sin2×30°/10

⇒ R = 1600 × sin60°/10

⇒ R = [1600 × √3/2]/10

⇒ R = 1385.6/10

⇒ R = 138.56 m

Answer 2

Answer:

Given :-

• A body is projected with initial velocity of 40 m/s and at an angle of 30° with the horizontal. (assume g = 10 m/s²).

To Find :-

1. Time of flight
2. Horizontal range of projectile.

Formula Used :-

$\clubsuit$ Time Taken :

$\longmapsto\sf\boxed{\bold{\pink{Time\: taken\: (t) =\: \dfrac{2vsin\theta}{g}}}}$

$\clubsuit$ Horizontal range :

$\longmapsto\sf\boxed{\bold{\pink{Horizontal\: range\: (R) =\: \dfrac{{v}^{2}sin2\theta}{g}}}}$

where,

• v = Velocity
• g = Acceleration due to gravity

Solution :-

${\small{\bold{\purple{\underline{i)\: Time\: of\: flight\: :-}}}}}$

Given :

• Velocity = 40 m/s
• Acceleration due to gravity = 10 m/s²
• Angle of projection = 30°

According to the question by using the formula we get,

$\implies \sf Time\: of\: flight =\: \dfrac{2 \times 40 \times sin30^{\circ}}{10}$

As we know that, [ sin30° = 1/2 ]

$\implies \sf Time\: of\: flight =\: \dfrac{\cancel{80} \times \dfrac{1}{\cancel{2}}}{10}$

$\implies \sf Time\: of\: flight =\: \dfrac{4\cancel{0}}{1\cancel{0}}$

$\implies \sf Time\: of\: flight =\: \dfrac{4}{1}$

$\implies\sf\bold{\red{Time\: of\: flight =\: 4\: seconds}}$

$\therefore$ The time of flight is 4 seconds.

$\rule{150}{2}$

${\small{\bold{\purple{\underline{ii)\: Horizontal\: range\: of\: projectile\: :-}}}}}$

Given :

• Velocity = 40 m/s
• Acceleration due to gravity = 10 m/s²
• Angle of projection = 30°

According to the question by using the formula we get,

$\implies \sf Horizontal\: Range =\: \dfrac{{(40)}^{2} \times sin2 \times 30^{\circ}}{10}$

$\implies \sf Horizontal\: Range =\: \dfrac{1600 \times sin60^{\circ}}{10}$

As we know that, [ sin60° = √3/2 ]

$\implies \sf Horizontal\: Range\: =\: \dfrac{\cancel{1600} \times \dfrac{\sqrt{3}}{\cancel{2}}}{10}$

$\implies \sf Horizontal\: Range =\: \dfrac{80\cancel{0} \times \sqrt{3}}{1\cancel{0}}$

$\implies \sf Horizontal\: Range =\: \dfrac{80 \times \sqrt{3}}{1}$

$\implies \sf Horizontal\: Range =\: 80 \times \sqrt{3}$

$\implies \sf\bold{\red{Horizontal\: Range =\: 138.56\: m}}$

$\therefore$ The horizontal range of projectile is 138.56 m .

$\rule{150}{2}$

EXTRA INFORMATION :

$\clubsuit$ Maximum Height :

$\longmapsto \sf \boxed{\bold{\pink{Maximum\: Height =\: \dfrac{{v}^{2}{sin}^{2}\theta}{2g}}}}$

where,

• v = Velocity
• g = Acceleration due to gravity