Question

During the battle of Bunker Hill, Colonel William Prescott ordered the American Army to bombard the British Army camped near Boston. The projectiles had an initial velocity of 41 m/s at 38° above the horizon and an initial position that was 35 m higher than where they hit the ground. How far did the projectiles move horizontally before they hit the ground? Ignore air resistance.

Answer 1

Step-by-step explanation:

initial velocity, u = 41 m/s

angle, θ = 38 °

height, h = 35 m

Let the time is t.

Use second equation of motion in vertical direction

h = ut + 1/2 gt²

- 35 = 41 Sin 38 t - 0.5 x 9.8 x t²

4.9t² - 25.2 t - 35 = 0

`t = \frac{25.2 \pm \sqrt{25.2^(2)+4* 4.9* 35}}{2* 4.9}`

t = 6.3 second

Horizontal distance traveled in time t is

d = uCos 38 x t

d = 41 x Cos 38 x 6.3

d = 203.54 m

Answer 2

`s=203.149\ m`

Step-by-step explanation:

Given:

• initial velocity of projectile,
  `u=41\ m.s^(-1)`

• angle of projection above horizontal,
  `\theta=38^(\circ)`

height of the initial projection point above the ground,
`y=35\ m`

Vertical component of the velocity:

`u_y=u.\sin\theta`

`u_y=41* \sin38`

`u_y=25.242\m.s^(-1)`

The time taken in course of going up:

(at top the final velocity will be zero)

`v_y=u_y-g.t`

`0=25.242-9.8* t`

`t=2.576\ s`

In course of going up the maximum height reached form the initial point:

(at top height the final velocity is zero. )

using eq. of motion,

`v_y^2=u_y^2-2* g.h`

where:

`v_y=` final vertical velocity while going up.=0

`h=` maximum height

`0^2=25.242^2-2* 9.8* h`

`h=32.5081\ m`

Now the total height to be descended:

`h'=h+y`

`h'=32.5081+35`

`h'=67.5081\ m`

Now the time taken to fall the gross height in course of falling from the top:

`h'=v_y.t'+(1)/(2) g.t'^2`

`67.5081=0+4.9* t'^2`

`t'=3.7118\ s`

Now the total time the projectile spends in the air:

`t_t=t+t'`

`t_t=2.576+3.7118`

`t_t=6.2878\ s`

Now the horizontal component of the initial velocity:

(it remains constant throughout the motion)

`u_x=u.\cos\theta`

`u_x=41* \cos38`

`u_x=32.3084\ m.s^(-1)`

Therefore the horizontal distance covered in the total time;

`s=u_x* t_t`

`s=32.3084* 6.2878`

`s=203.149\ m`