16.0-kg cannonball is fired from a cannon with muzzle speed of 800 m/s at an ang

Question

16.0-kg cannonball is fired from a cannon with muzzle speed of 800 m/s at an angle of 32.0 degree with the horizontal. A second ball is fired with the same initial speed at an angle of 90.0 degree. Let y = 0 at the cannon. (a) Use the isolated system model to find the maximum height reached by each ball. = 12918.87 h^first ball do not try to use equations involving projectile motion. This problem is much more straightforward to solve using energy conservation, m h^second _ m ball Use the isolated system model to find the total mechanical energy of the ball-Earth system at the maximum height for each ball. E_first ball = J E^second ball = J

Explanation / Answer

given

muzzle speed of first ball =800 m/sec

first ball shot at an angle = 32 degrees

now maximum height reached by first ball = v^2sin^2teta/2g

A) H1 = 800*800*sin^2(32degrees)/2*9.8

H1= 1.80048* 10000= 18004.8m

height reached by second ball = u^2/2g because sin90 =1

H2 = 800*800/19.6= 3.265*10000m

b) now mechanical energy of first ball is

E1= mgH1 = 16*9.8*18004.8= 2823152.64

E2= mgH2 = 16*9.8* 3.265*10000 = 511.952 * 10000.

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