Answer :
Answer:
a) t = 5.40 s
b) [tex]E_{p} = 684.3 J[/tex]
c) [tex] E_{k} = 1559.52 J [/tex]
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:
[tex]-20 = v_{0}sin(\theta)*t - \frac{1}{2}gt^{2}[/tex] (1)
Where g is the gravity = 9.81 g/s²
By solving equation (1) for t, we have:
[tex] t = 5.40 s [/tex]
b) The gravitational potential energy ([tex]E_{p}[/tex]) of the projectile at point P (at the maximum height in the parabolic motion) is the following:
[tex] E_{p} = mgh_{P} [/tex]
Where [tex]h_{P}[/tex]: is the height at the point P. This can be calculated using the following equation:
[tex]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[/tex]
Now, the gravitational potential energy is:
[tex]E_{p} = mgh_{P} = 1.5 kg*9.81 m/s^{2}*46.5 m = 684.3 J[/tex]
c) The kinetic energy of the projectile at point R (the same height as the edge of the building in the parabolic motion) is:
[tex] E_{k} = \frac{m*v_{R}^{2}}{2} [/tex]
Where [tex]v_{R}[/tex] is the velocity at the point R, which is:
[tex]v_{R} = -45.6 m/s[/tex]
Now, the kinetic energy is:
[tex] E_{k} = \frac{m*v_{R}^{2}}{2} = \frac{1.5 kg*(-45.6 m/s)^{2}}{2} = 1559.52 J [/tex]
d) The magnitude and direction of impulse the projectile impacts on the ground can be calculated using the equation:
[tex]F*t = m*v_{Q}[/tex]
Where F: is the force
The velocity at the point Q is:
[tex]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[/tex]
Hence, the magnitude and direction of impulse is:
[tex]F*t = m*v_{Q} = 1.5 kg*49.7 m/s = 74.6 kgm/s[/tex]
I hope it helps you!