Question

An animal-rescue plane flying due east at 49 m/s drops a bale of hay from an altitude of 66 m . The acceleration due to gravity is 9.81 m/s 2 . If the bale of hay weighs 160 N , what is the momentum of the bale the moment it strikes the ground? Answer in units of kg m/s.At what angle of inclination will the bale strike?

Answer 1

Explanation:

Horizontal velocity  = 49 m/s

Altitude, S = 66m

g= 9.81 m/s^2

weight W = 160N

mass m = w/g = 160/ 981 = 16.309 kg

when a object dropped from time time to reach the ground can be calculated from this equation.

S = Vot + 1/2 at^2

V_0 is the initial velocity in vertical direction. The plane moving horizontally an altitude of 66m it has only horizontal velocity.

The velocity V_0 in vertical direction is zero

S= 1/2 at^2

66=1/2 9.81 × t^2

t^2 = 13.455

t= 3.668 seconds                                 (Time to hit the hay in ground)

Vertical  velocity V = V _c + at            (v_0 is zero, initial vertical velocity )

V = gt                                                    (acceleration is 'g')

V = 9.81 × 3.668

V = 35.98 m/s

The hay has two components of velocity when it reach the ground the horizontal velocity 49 m/s it will  not change vertical velocity of 35.98 m/s

Resultant velocity can found by vector addition

velocity = $\sqrt{(49)^2 + (35.48)^2}$

velocity = 60.78 m/s

The momentum of bale of hay = mv

= 16.309 × 60.78

The momentum of bale of hay =991.26 kg m/s

Answer 2

Explanation:

The final velocity of the bale have vertical and horizontal components. So, we have:

$v^2=v_x^2+v_y^2(1)$

According to conservation of energy:

$\frac{mv_y^2}{2}=mgh\\v_y=\sqrt{2gh}$

Since the plane is flying due east, we have $v_x=49\frac{m}{s}$. Replacing in (1):

$v=\sqrt{(\sqrt{2gh})^2+v_x^2}\\v=\sqrt{2(9,81\frac{m}{s^2})(66m)+(49\frac{m}{s})^2}\\v=60.78\frac{m}{s}$

Now, we calculate the momentum of the bale:

$p=mv\\p=\frac{160N}{9.81\frac{m}{s^2}}(60.78\frac{m}{s})\\p=991.32\frac{kg\cdot m}{s}$

From pythagoras theorem we know that:

$tan\theta=\frac{opposite}{adjacent}\\tan\theta=\frac{v_x}{v_y}\\\theta=tan^{-1}(\frac{v_x}{v_y})\\\theta=tan^{-1}(\frac{60,78\frac{m}{s}}{49\frac{m}{s}})\\\theta=51.12^\circ$