a cannonball shot from a buried Cannon is fired at an angle of 37 degrees above

Question

a cannonball shot from a buried Cannon is fired at an angle of 37 degrees above a level field. The initial velocity has a magnitude of 50 m/s. In the questions that follow, you may neglect air resistance and assume that gravity equals 10 meters per second squared.

A. find the horizontal and vertical components of the vector VI
B. how many seconds does the Cannonball stay in the air?
C. how far from the cannon will the ball land?
D. how high above the field is the cannonball at its highest point?

Explanation / Answer

A .v=50 m/s

horizontal component=50cos(37)=39.93 m/s

vertical component=50sin(37)=30.09 m/s

B

time of flight=2vsin(37)/g=6.14 sec

C

horizontal range=v^2*sin(2*37)/g=245.2 m

it will land 245.2 m from cannon

D

maximum height=v^2sin^2(37)/2g=46.2 m

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