More expensive & probably less complicated
Answer:
Explanation:
Check attachment for solution
Let the cannonball be thrown at a height of h above ground.
Then the potential energy of the ball is
V = m*g*h
where
m = the mass of the ball
g = 9.8 m/s²
Also, the kinetic energy of the ball is
K = (1/2)mu²
where
u = 5 m/s, the vertical launch velocity.
Ignore wind resistance.
Because the total energy is preserved, the total energy (n the form of only kinetic energy) when the ball strikes the ground is
(1/2)mV²
where V = vertical velocity when the ball strikes the ground.
Expressions for both the initial and final energy are equal regardless of whether the ball s thrown downward or upward.
Therefore there is no difference in the landing speed.
Answer: There is no difference.
Well, <span>v = u + a×t is the equation.</span>
<span>
v: final velocity, which is 23 m/s in this equation.</span>
<span>u: initialo velocity = 13 m/s </span>
<span>a: acceleration = ? </span>
<span>t: time = 30s
</span>
Your equation would be...
<span>23 = 13 + a×30 </span>
<span>a = (23 - 13) / 30 </span>
<span>a = 1 / 3 </span>
<span>a = 0.333 m/s</span>
Two things are said to be in contact if the smallest distance between a point in one of them and a point in the other one is zero.
