Question

A body is projected with a velocity of 30m/s to have a horizontal range of 45m

Answer 1

Answer:

The body must be projected at an angle of 14.67° to achieve the desired output.

Explanation:

Here,

Initial velocity, u= 30m/s

Horizontal range, R= 45m

Angle of projection, n=?

We know, Horizontal range,

R=(u^2 sin2n)/g

or, sin2n= Rg/u^2

or, sin2n= (45×9.8)/(30)^2

or, sin2n= 0.49

or, 2n= sin^-1(0.49)

or, 2n= 29.34

or, n= 14.67

So, the body has to be projected at an angle of 14.67° to acquire the desired horizontal range at the selected velocity.

Answer 2

A body is projected with a velocity of 30m/s to have a horizontal range of 45m. then we have to find angle of projection of the body.

Let angle of projection is α.

then, horizontal range of body, R = u²sin2α/g

where , u is initial velocity of body, g is acceleration due to gravity.

here, R = 45m, u = 30m/s , g = 10m/s²[ experimental value ]

so, 45m = (30m/s)²sin2α/(10m/s²)

or, 45 × 10 = 900 sin2α

or, sin2α = 1/2 = sin30° or sin150°

or, 2α = 30° or 150°

or, α = 15° or 75°

hence, angle of projection is either 15° or 75°.