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

Answer:

$s=203.149\ m$

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:

$u_y=u.\sin\theta$

$u_y=41\times \sin38$

$u_y=25.242\m.s^{-1}$

The time taken in course of going up:

(at top the final velocity will be zero)

$v_y=u_y-g.t$

$0=25.242-9.8\times t$

$t=2.576\ s$

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,

$v_y^2=u_y^2-2\times g.h$

where:

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

$h=$ maximum height

$0^2=25.242^2-2\times 9.8\times h$

$h=32.5081\ m$

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:

$h'=v_y.t'+\frac{1}{2} g.t'^2$

$67.5081=0+4.9\times t'^2$

$t'=3.7118\ s$

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)

$u_x=u.\cos\theta$

$u_x=41\times \cos38$

$u_x=32.3084\ m.s^{-1}$

Therefore the horizontal distance covered in the total time;

$s=u_x\times t_t$

$s=32.3084\times 6.2878$

$s=203.149\ m$

Answer 2

Answer:

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

$t = \frac{25.2 \pm \sqrt{25.2^{2}+4\times 4.9\times 35}}{2\times 4.9}$

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