Question

A body is projected upward making angle Thita with the horizontal with a velocity of 300 m/s. Find the value of Thita so that the horizontal range will be maximum. Hence find its range and time of flight




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Answer 1

To find the value of theta lets the initial velocity and acceleration due to gravity be constant.

Now by formula we know that,

R = u²Sin 2Ø/g

R ∞ Sin 2Ø

This shows that range will be maximum when SinØ will be maximum and we know that maximum value of SinØ = 1

Sin 2(Ø) = 1

2Ø = 90°

Ø = 45°

Hence, range will be maximum when the projection with horizontal = Ø = 45°

Now, putting the given Values to find the maximum range.

R = u²Sin2Ø/g

R = 300 × 300 × Sin(90°)/10

$\bf{{R}_{max}\:=\:9000\:m}$

Again,

We know that

$R \:=\:{u}_{x}T$

9000 = 300 × T

$\bf{T\:=\:30\:s}$

Answer 2

Given :-

A body is projected upward making angle Thita with the horizontal with a velocity of 300 m/s.

To Find :-

Range

Time of flight

Solution :-

Sin 2$\theta$ = 1

2Ø = 90°

Ø = 90/2

Ø = 45°

Now

Maximum height = u² sin²Ø/g

Maximum height = (300)² × sin(90)/10

Maximum height = 90000 × 1/10

Maximum height = 9000 m

Now

Time = 60 × sin45

Time  = 60 × √1/2

Time = √60/2

Time = 42.4 sec