Question

A ball thrown into space with a velocity of 20m/s at an angle of 45° with respect to the horizontal. calculate
1. maximum height
2.time of flight
3.horizontal range ( given g =9.8m/s^2)​

Answer 1

Given :-

• Velocity (u) = 20 m/s
• Angle (θ) = 45°
• Acceleration due to gravity (g) = 9.8 m/s²

To Find :-

1. Maximum height
2. Time of flight
3. Horizontal range

Solution :-

Case (I), Maximum height

As we know that,

★ h = u²sin²θ/2g

[ Put the values ]

⟶ h = 20² × sin²(45°)/2 × 9.8

⟶ h = 400 × (1/√2)²/19.6

⟶ h = 400 × 1/2 /19.6

⟶ h = 200 × 19.6

⟶ h = 3920 m ★

Case (II), Time of flight

We know that,

★ T = 2u sinθ/g

[ Put the values ]

⟶ T = 2(20) × sin45°/9.8

⟶ T = 40 × 1/√2 /9.8

⟶ T = 40/√2 / 49/5

⟶ T = 40 × 5 / 49√2

⟶ T = 200/49√2

⟶ T = 200 / 49 × 1.732

⟶ T = 200/84.868

⟶ T = 2.3 sec ★

Case (III), Horizontal range

We know,

★ R = u² sin2θ/g

[ Put the values ]

⟶ R = (20)² × sin2(45°)/9.8

⟶ R = 400 × sin90°/9.8

⟶ R = 400 × 1 /9.8

⟶ R = 400/9.8

⟶ R = 40.8 m ★

Hence, solved !!