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Question

1. An object at an angle such that the horizontal range is 4 times of the maximum height .what is the angle projection of the object?

2. Football player hits the ball with speed 20 m per second with an angle 30 degree with respect too original direction .the goal post is a distance of 40 m from him. find out whether the ball reach the goal post.

3. If an object is thrown horizontally with an initial speed 10 M per second from the top of the building of height hundred metre. what is horizontal distance covered by the particle?

4. An object is thrown with initial speed 5 M per second set an angle of projection 30 .what is the height and range reached by the particle?

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Solution

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ask one question at a time

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Final answer : Angle of Projection is 45°

STEPS:

1) According to the question,

Horizontal Range = 4 * (Maximum Height) ---(1)

2) We know that when prijectile is fired at angle ( a) ,then

H(max) = u^2( sin^2(a)) / 2g

Horizontal Range, R = u^2 sin(2a)/g

where u is initial velocity.

Using equation (1),

u^2 sin(2a)/g = 4 * u^2sin^2a/2g

===> sin(2a)=2sin^2 (a)

===>2sina cos a =2sin^2 a

===>sina =cos a

====>a=15

hence a is not equal to 0

hence projection angle is 45 degree

3) This is the case of horizontal projectile.

Given, initial velocity , u = 20 m/s

height , h = 100m

use the formula,

Horizontal range , X = u√(2h/g)

So, X = 20 × √{2 × 100/10}

X = 20 × √20 m

Hence, horizontal distance covered by object is 20√20 m

ask one question at a time

now i will answer 2 question for your sake

Final answer : Angle of Projection is 45°

STEPS:

1) According to the question,

Horizontal Range = 4 * (Maximum Height) ---(1)

2) We know that when prijectile is fired at angle ( a) ,then

H(max) = u^2( sin^2(a)) / 2g

Horizontal Range, R = u^2 sin(2a)/g

where u is initial velocity.

Using equation (1),

u^2 sin(2a)/g = 4 * u^2sin^2a/2g

===> sin(2a)=2sin^2 (a)

===>2sina cos a =2sin^2 a

===>sina =cos a

====>a=15

hence a is not equal to 0

hence projection angle is 45 degree

3) This is the case of horizontal projectile.

Given, initial velocity , u = 20 m/s

height , h = 100m

use the formula,

Horizontal range , X = u√(2h/g)

So, X = 20 × √{2 × 100/10}

X = 20 × √20 m

Hence, horizontal distance covered by object is 20√20 m

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