Question

A 20.0-kg cannonball is fired from a cannon with muzzle speed of 1 000 m/s at an angle of 37.08 with the horizontal. A second ball is fired at an angle of 90.08. Use the isolated system model to find (a) the maximum height reached by each ball and (b) the total mechanical energy of the ball–Earth system at the maximum height for each ball. Let y5 0 at the cannon.

Answer 1

Answer:

(a). The maximum height by first ball is $1.8545\times10^{4}\ m$

The maximum height by second ball is $5.1020\times10^{4}\ m$

(b). The total mechanical energy of the ball–Earth system at the maximum height for each ball is $1.0\times10^{7}\ J$

Explanation:

Given that,

Mass of cannonball = 20.0 kg

Speed = 1000 m/s

Angle with horizontal= 37.08

Fired angle = 90.08

We need to calculate the speed of the ball

Using formula of speed

$v_{y}=v\sin\theta_{H}$

Put the value into the formula

$v_{y}=1000\times\sin37.08$

$v_{y}=602.9\ m/s$

(a). We need to calculate the maximum height by first ball

Using conservation of energy

$\dfrac{1}{2}mv^2=mgh$

$h= \dfrac{v^2}{2g}$

Put the value into the formula

$h=\dfrac{(602.9)^2}{2\times9.8}$

$h=1.8545\times10^{4}\ m$

We need to calculate the maximum height by second ball

Using conservation of energy

$\dfrac{1}{2}mv^2=mgh$

$h= \dfrac{v^2}{2g}$

Put the value into the formula

$h=\dfrac{(1000)^2}{2\times9.8}$

$h=5.1020\times10^{4}\ m$

(b). We need to calculate the total mechanical energy of the ball–Earth system at the maximum height for each ball

Using formula of energy

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times20\times1000^2$

$E=1.0\times10^{7}\ J$

Hence, (a). The maximum height by first ball is $1.8545\times10^{4}\ m$

The maximum height by second ball is $5.1020\times10^{4}\ m$

(b). The total mechanical energy of the ball–Earth system at the maximum height for each ball is $1.0\times10^{7}\ J$