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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.

2 Answers

3 votes

Answer:

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

User Nerian
by
5.3k points
3 votes

Answer:


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

User Rahul Premraj
by
5.8k points