A projectile is fired with an initial speed of 37.0 m/s at an angle of 44.3 ∘ above the horizontal on a long flat firing range.

Question

Find:

A. Determine the maximum height reached by the projectile.

B. Determine the total time in the air.

C. Determine the total horizontal distance covered (that is, the range).

D. Determine the speed of the projectile 1.00 after firing.

Answer 1

Hello Bri

H max = Max Height attained

T = time of Full flight

R= u² Sin 2 theta/g = Horizontal Range

Vt=Speed at time t

Let us start with out knowing a formula

The initial speed u has two components

Horizontal component of speed = uCos theta

and vertical speed component = uSin theta

Here Horizontal component remains unchanged al through the flight , whereas vertical component is has gravity influence

For Horizontal motion Let T= total time of fight

So Speed = Distance/Time = R/T, where R = range

uCos theta = R/T

For vertical motion, Initial speed =uSin theta

Final speed at height point = 0

Using V² =U² -2gH

As V=0 , (uSin theta) ² = 2gH

So H max = u²Sin²theta/2g Plug in u= 37, theta =43.3

Get H max

Fot time up to highest point

v = u-gt

, Where , v= 0

uSintheta =gt

So t= Time to cover half flight = uSintheta /g

Plug in again u =37 and theta= 43.3

Get t and T = 2t = Time it remains in air= Time of full flight

Thus range =(u Cos theta) 2t= uCos theta ( 2usintheta/g) = u² Sin(2Theta)/g

For speed at given time

Horizontal component is constant = uCos theta and vertical speed at given time = 1 sec is

v= uCos theta-g as t= 1 sec

So resultan√t speed a 1 sec = √( uCos teta -g)² +( uSin theta)²

Again use u = 37 , theta =43.3 , g = 9.8