Question

A body is thrown with the velocity v at an angle alpha with the horizontal. If the body remains in air for 6

seconds, the maximum height reached by the body will beA body is thrown with the velocity vo at an angle  with the horizontal. If the body remains in air for 6

seconds, the maximum height reached by the body will be​

Answer 1

Answer:

Explanation:

Let the body is projected with velocity u and the angle of projection be θ .

Given : Time of flight     T=4 s

∴  Time of flight         T=  

g

2usinθ

​  

 

⟹       u  

2

 sin  

2

θ=  

4

T  

2

g  

2

 

​  

           .......(1)

Maximum height reached         h=  

2g

u  

2

sin  

2

θ

​  

=  

2g

4

T  

2

g  

2

 

​  

 

​  

=  

8

T  

2

g

​  

             

∴    h=  

8

4  

2

×10

​  

=20 m

Answer 2

Concept:

Projectile motion: An item or particle that is propelled toward the surface of the Earth and moves along a curved route only under the influence of gravity is said to be in projectile motion.

To find: The maximum height reached by the object.

Given: The object is thrown with a velocity  'V' at an angle alpha with the horizontal line.

Explanation:

Total time taken that is,

T=    2$usin\alpha$/g  = 6

    ∴  $usin\alpha$ = 3g

The maximum height reached that is,

H =   $u^2sin\alpha /2g$

   =   $9g^2/2g$ = 4.5g = 44.1m

where 'g' refers to acceleration due to gravity.

        Answer: 44.1 m

So when an object is thrown at a velocity v at an angle alpha with the ground or horizontal line, then the maximum height reached by the object will be 44.1 m.

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