Question

A 21.0-kg cannonball is fired from a cannon with muzzle speed of 950 m/s at an angle of 33.0° with the horizontal. A second ball is fired with the same initial speed at an angle of 90.0°. Let y = 0 at the cannon.
(a) Use the isolated system model to find the maximum height reached by each ball. hfirst ball = Incorrect: Your answer is incorrect. Do not try to use equations involving projectile motion. This problem is much more straightforward to solve using energy conservation. m hsecond ball =What is the expression for the total mechanical energy just after the cannon has been fired? What is the expression for the total mechanical energy when the cannonball reaches its maximum height? m
(b) Use the isolated system model to find the total mechanical energy of the ball❝Earth system at the maximum height for each ball. Efirst ball = Incorrect: Your answer is incorrect. How does the total mechanical energy at the maximum height compare to the total mechanical energy immediately after the cannon was fired? J Esecond ball =

Answer 1

Answer:

v2f1−v2i1=2gh1

Where, vf1 is the final velocity of first ball, vi1 is the initial velocity of first ball and h1 is the maximum height.

Now we substitute 00 for vf1, 1000 m/s(sin37°)1000 m/s(sin37°) for vi1and −9.81 m/s2−9.81 m/s2 for gg to find h1 which is the height attained.

= 0^2−(1000 m/s(sin37°))^2=2(−9.81 m/s2)h1

Now we divide through

h1=1m

Now the mechanical energy is determined using.

E = mgh

= 21 ×9.8 × 1

= 205.8J