A cannon sends a projectile towards a target a distance 1340 maway. The initial

Question

A cannon sends a projectile towards a target a distance 1340 maway. The initial velocity makes an angle 33 degrees with thehorizontal. The target is hit The acceleration of gravity is 9.8 m/s2. Whatis the magnitude of the initial velocity? Answer in units ofm/s. How high is the highest point of the trajectory? Answerin units of m. How long does it take for the projectile to reach thetarget? (Assume no friction) answer in units ofs. Please show work...I'm having a hard time with this one. A cannon sends a projectile towards a target a distance 1340 maway. The initial velocity makes an angle 33 degrees with thehorizontal. The target is hit The acceleration of gravity is 9.8 m/s2. Whatis the magnitude of the initial velocity? Answer in units ofm/s. How high is the highest point of the trajectory? Answerin units of m. How long does it take for the projectile to reach thetarget? (Assume no friction) answer in units ofs. Please show work...I'm having a hard time with this one.

Explanation / Answer

For PART A using the projectile motion formula d= (V0^2 *sin2)/g V0= (d*g)/Sin 2 =(1340)(9.8)/sin 2(33) = 120.13m/sec. I will be working on B and C. we know that at the highest point velocity =0, therefore Vfy=Viy-gt t=(-120.13)/(-9.8) = 12.3 seconds therefore highest point is h=1/2 * g*t^2 = 1/2 (9.8)(12.3)^2 =736.3 m time taken to reach the target is t=d/v = (1340)/120 = 11.16seconds Note expect fot the first answer , the rest i am not sure. I amjust learning physics

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