A cannon sends a projectile towards a target a distance 1350 maway. The initial

Question

A cannon sends a projectile towards a target a distance 1350 maway. The initial velocity makes an angle of 23 degrees withthe horizontal. The target is hit. The aceleration of gravity is 9.8 meters per second. How high is the highest point of the trajectory? A cannon sends a projectile towards a target a distance 1350 maway. The initial velocity makes an angle of 23 degrees withthe horizontal. The target is hit. The aceleration of gravity is 9.8 meters per second. How high is the highest point of the trajectory?

Explanation / Answer

let initial velocity=u initial horizontal velocity=u*cos23 horizontal acceleration=0 1350=u*cos23*t initial vertical velocity=u*sin23 vertical acceleration=g=9.8 at highest point the projectile's vertical velocity=0 h=height of the highest point t/2 is the time taken to reach the highest point v=u-g*t 0=u*sin23-9.8*(t/2) u*sin23=4.9*t (u*sin23)/(u*cos23*t)=(4.9*t)/1350 1350*tan23=4.9*t2 t=10.81 u*sin23=4.9*10.81 u=135.81 m/s v2=u2-2*g*h 0=(135.81*sin23)2-2*9.8*h h=143.67 m highest point of the trajectory=143.67 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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