Question

A projectile is thrown in vertical plane with speed 42 m/s at an angle of 45° with horizontal surface. Range of this projectile is [g = 10 m/s-)​

Answer 1

Answer :

• R = 176.4 m

S O L U T I O N :

Given,

• Initial velocity (u) = 42 m/s
• Angle (θ) = 45°
• Acceleration due to gravity (g) = 10 m/s²

To Find,

• The range of this projectile.

Explanation,

We know that,

R = u²sin2θ/g

[ Put the values ]

=> R = (42)² × sin2(45°)/10

=> R = 1764 × sin90°/10

[ °.° sin90° = 1 ]

=> R = 1764 × 1/10

=> R = 1764/10

=> R = 176.4 m

Therefore,

The range of this projectile is 176.4 m.

Answer 2

Answer:

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