2. 10 points SerPSE9 8.P.004.W. My Notes Ask Your Teacher A 21.0-kg cannonball i

Question

2. 10 points SerPSE9 8.P.004.W. My Notes Ask Your Teacher A 21.0-kg cannonball is fired from a cannon with muzzle speed of 900 m/s at an angle of 36.0° 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 hsecond ball = (b) Use the isolated system model to find the total mechanical energy of the ball-Earth system at the maximum height for each ball Ernst ball = Esecond ball

Explanation / Answer

(a) <Cannonball at 36 degrees>

We need the equation of motion that contain v, u, a & s because we know atleast 3 of these variables.

This is it:
v2 = u2 + 2as


Initial vertical velocity (u) of the ball is 900sin36 = 529.0067 m/s upwards
a = 9.81 m/s2 downwards

At maximum height vertical velocity v = 0 m/s
v2 = u2 + 2as
0 = 529.00672 + 2*-9.81*s
s = 14263.41067 m


<90 degrees>

u = 900 m/s
v = 0 m/s
a = -9.81 m/s^2

v2 = u2 + 2as
0 = 9002 + 2*-9.81*s
s = 41284.40367 m


(b)  

<37 degrees>
KE = (1/2)*m*v2
KE = (1/2)*21*(900cos36)2 << at maximum height vertical velocity is zero, so the resultant velocity is only the horizontal part.

=5566594.769J
And potential energy will be ,mgh

So,

21*9.81*14263.41067

2938405.232J

total energy= 5566594.769+2938405.232

==8505000J

<90 degrees>
there will be only potential energy

SO,

mgh

21*9.81*41284.40367

=8505000J

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