Question

g A 1.5 kg projectile is fired from the edge of a 20-m tall building with an initial velocity of v0 = 45.6 m/s directed 30° above horizontal. The projectile rises to point P and then lands on ground at Q. (a) Find the flight time of the projectile as it lands at Q. (b) Calculate gravitational potential energy of the projectile at point P. (c) Compute kinetic energy of the projectile at point R, which is at the same height as the edge of the building. (d) Determine the magnitude and direction of impulse the projectile impacts on the ground.

Answer 1

Answer:

a) t = 5.40 s

b) $E_{p} = 684.3 J$

c) $E_{k} = 1559.52 J$

d) mv = 74.6 kgm/s

Explanation:

We have:

v₀ : initial speed = 45.6 m/s

θ: 30 °

h: height = 20 m

m: projectile's mass = 1.5 kg  

     

a) The flight time of the projectile as it lands at Q (when it impacts on the ground in the parabolic motion) can be calculated using the following equation:          

$-20 = v_{0}sin(\theta)*t - \frac{1}{2}gt^{2}$   (1)

Where g is the gravity = 9.81 g/s²

By solving equation (1) for t, we have:

$t = 5.40 s$

b) The gravitational potential energy ($E_{p}$) of the projectile at point P (at the maximum height in the parabolic motion) is the following:  

$E_{p} = mgh_{P}$

Where $h_{P}$: is the height at the point P. This can be calculated using the following equation:

$h_{P} = h + \frac{v_{0}^{2}(sin(\theta))^{2}}{2g} = 20 m + \frac{(45.6 m/s)^{2}(sin(30))^{2}}{2*9.81 m/s^{2}} = 46.5 m$        

Now, the gravitational potential energy is:

$E_{p} = mgh_{P} = 1.5 kg*9.81 m/s^{2}*46.5 m = 684.3 J$  

c) The kinetic energy of the projectile at point R (the same height as the edge of the building in the parabolic motion) is:

$E_{k} = \frac{m*v_{R}^{2}}{2}$          

Where $v_{R}$ is the velocity at the point R, which is:

$v_{R} = -45.6 m/s$      

Now, the kinetic energy is:

$E_{k} = \frac{m*v_{R}^{2}}{2} = \frac{1.5 kg*(-45.6 m/s)^{2}}{2} = 1559.52 J$          

d) The magnitude and direction of impulse the projectile impacts on the ground can be calculated using the equation:

$F*t = m*v_{Q}$

Where F: is the force

          

The velocity at the point Q is:

$v_{Q} = \sqrt{(v_{q_{y}})^{2} + (v_{q_{x}}})^{2}} = \sqrt{(-30.17 m/s)^{2} + (39.49 m/s)^{2}} = 49.7 m/s$      

Hence, the magnitude and direction of impulse is:                                              

$F*t = m*v_{Q} = 1.5 kg*49.7 m/s = 74.6 kgm/s$    

I hope it helps you!