Question

In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of your rich uncle. The top of the fence is 5.00 m above the ground. You have taped an important message to a rock that you want to throw over the fence. The ground is level, and the width of the fence is small enough to be ignored. You throw the rock from a height of 1.60 m above the ground and at an angle of 56.0o above the horizontal. (a) What minimum initial speed must the rock have as it leaves your hand to clear the top of the fence? (b) For the initial velocity calculated in part (a), what horizontal distance be-yond the fence will the rock land on the ground?

Answer 1

Answer:

a) $u=13.3032\ m.s^{-1}$

b) $s=3.7597\ m$

Explanation:

Given:

horizontal distance between the fence and the point of throwing, $r=14\ m$

height of the fence from the ground, $h_f=5\ m$

height of projecting the throw above the ground, $h'=1.6\ m$

angle of projection of throw from the horizontal, $\theta=56^{\circ}$

• Let the minimum initial speed of projection of the throw be u meters per second so that it clears the top of the fence.
• Now the effective target height, $h=h_f-h'=5-1.6=3.4\ m$

The horizontal component of the velocity that remains constant throughout the motion:

$u_x=u\cos\theta$

Now the time taken to reach the distance of the fence:

use equation of motion,

$t_f=\frac{r}{u_x}$

$t_f=\frac{14}{u.\cos56}$ .................................(1)

Now the time taken to reach the fence height (this height must be attained on the event of descending motion of the rock for the velocity to be minimum).

Maximum Height of the projectile:

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

$h_m=\frac{u_y^2}{19.6}$ ........................(4)

Now the height descended form the maximum height to reach the top of the fence:

$\Delta h=h_m-h'$

$\Delta h=(\frac{u_y^2}{19.6} -3.4)\ m$

time taken to descent this height from the top height:

$\Delta h=u_{yt}.t_d+\frac{1}{2} \times g.t_d^2$

where:

$u_{yt}=$  initial vertical velocity at the top point

$t_d=$ time of descend

$(\frac{u_y^2}{19.6} -3.4)=0+0.5\times 9.8\times t_d^2$

$t_d=\sqrt{(\frac{u_y^2}{96.04} -\frac{3.4}{4.9} )}$..............................(2)

So we find the time taken by the rock to reach the top of projectile where the vertical velocity is zero:

$v_y=u_y-g.t_t$

where:

$u_y=$ initial vertical velocity

$v_y=$ final vertical velocity

$t_t=$ time taken to reach the top height of the projectile

$0=u_y-g.t_t$

$t_t=\frac{u_y}{9.8}\ seconds$ .................................(3)

Now the combined events of vertical and horizontal direction must take at the same time as the projectile is thrown:

So,

$t_f=t_t+t_d$

$\frac{14}{u.\cos56}=\frac{u_y}{9.8} +\sqrt{(\frac{u_y^2}{96.04} -\frac{3.4}{4.9} )}$

$\frac{14}{u.\cos56}=\frac{u\sin56}{9.8} +\sqrt{(\frac{(u.\sin56)^2}{96.04} -\frac{3.4}{4.9} )}$

$\frac{196}{u^2.\cos^2 56} +\frac{u^2\sin^2 56}{96.04} -2.857\times \tan56=\frac{u^2\sin^2 56}{96.04} -0.694$

$u=13.3032\ m.s^{-1}$

Max height:

$h_m=\frac{(u.\sin 56)^2}{19.6}$

$h_m=\frac{(13.3032\times \sin56)^2}{19.6}$

$h_m=6.2059\ m$

Now the rock hits down the ground 1.6 meters below the level of throw.

Time taken by the rock to fall the gross height $h_g=h_m+h'$:

$h_g=u_{yt}.t_g+\frac{1}{2} g.t_g^2$

$7.8059=0+0.5\times 9.8\times t_g^2$

$t_g=1.2621\ s$

Time taken to reach the the top of the fence from the top, using eq. (2):

$t_d=\sqrt{(\frac{u_y^2}{96.04} -\frac{3.4}{4.9} )}$

$t_d=\sqrt{(\frac{(u.\sin56)^2}{96.04} -\frac{3.4}{4.9} )}$

$t_d=0.7567\ s$

Time difference between falling from top height and the time taken to reach the top of fence:

$\Delta t=t_g-t_d$

$\Delta t=1.2621-0.7567$

$\Delta t=0.5054\ s$

b)

Now the horizontal distance covered in this time:

$s=u.\cos56\times\Delta t$

$s=13.3032\times \cos56\times 0.5054$

$s=3.7597\ m$ is the horizontal distance covered after crossing the fence.

Answer 2

Answer:

Explanation:

a ) Height to be cleared = 5 - 1.6 = 3.4 m

Horizontal distance to be cleared = 5 m .

angle of throw = 56°

here y = 3.4 , x = 5 , θ = 56

equation of trajectory

y = x tanθ - 1/2 g ( x/ucosθ)²

3.4 = 5 tan56 - 1/2 g ( 5/ucos56)²

3.4 = 7.4 - 122.5 / .3125u²

122.5 / .3125u² = 4

u² = 98

u = 9.9 m /s

Range = u² sin 2 x 56 / g

= 9.9 x 9.9 x .927 / 9.8

= 9.27 m

horizontal distance be-yond the fence will the rock land on the ground

= 9.27 - 5

= 4.27 m