A projectile of mass 0.298 kg is shot from a cannon, at height 6.8 m, as shown i

Question

A projectile of mass 0.298 kg is shot from a cannon, at height 6.8 m, as shown in the figure, with an initial velocity vi having a horizontal component of 6 m/s. The projectile rises to a maximum height of
y above the end of the cannon

Explanation / Answer

Picture Not Given.. So i assume Angle is 51 degrees Given the angle of the trajectory and the horizontal component of muzzle velocity, we can determine the vertical component of muzzle velocity Vxo = 6ms = Vo*cos51 => 6/cos51 = Vo = 9.54m/s => Vyo= 9.54sin51=7.41m/s The peak altitude occurs when Vy= Vyo -gt = 0 => t = Vyo/g = 0.76s How high did it go? Height = 6.8+Vyo*t-0.5gt^2 = 6.8+7.41*0.76-4.9*0.76^2= 9.63 m delta y = 9.63-6.8=2.83m How long does it take to fall from this height? 9.63 = gt^2 => t = sqrt(9.63/9.8) = 0.99 = 1 s What is the vertical velocity upon impact Vy=gt = 9.8*1.0 = 9.8ms The horizontal velocity is 6 The magnitude of the impact velocity is sqrt(6^2+9.8^2)=11.49m/s magnitude of the velocity =11.49m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.