A stone mass M thrown straight up with initial velocity V reaches a height H. A second stone of mass 2M, thrown straight up with an initial velocity 2V. What will be its height?
for inital velocity v, we have
h(t) = vt-(g/2)t^2
This reaches a max height at t=v/g, and
h(v/g) = v(v/g) - (g/2)(v/g)^2
= v^2/g - v^2/2g = v^2/2g
So, we see that max height is directly proportional to v^2
So, double v, and you quadruple the max height.
M does not matter, as Galileo discovered.
according to the formula H=Vo*t-(gt^2)/2;
we can see that mass doesn't play any role here, so we can fint out our max. heights:
H2=2v*t2-(gt2^2)/2, and H1=v*t1-(gt1^2)/2;
Also t=v0/g => t1=v/g; t2=2v/g =>
H2=4v^2/g - 2v^2/g; H1=v^2/g-v^2/2g =>
H2=v^2(4/g-2/g); H1=v^2(1/g-1/2g) =>
H2/H1=4 => H2=4*H1. (I suggest to check my solution, mb I have mistakes)
To find the height reached by the second stone, we can make use of the principles of projectile motion.
First, let's analyze the motion of the first stone, which has a mass of M and is thrown straight up with an initial velocity of V. As the stone moves upward against the force of gravity, it slows down until it eventually reaches its peak height (H). At this point, the stone briefly comes to a halt before falling back down due to gravity.
Now, let's consider the second stone, which is twice as massive (2M) and thrown with twice the initial velocity (2V). Since the second stone has greater mass and initial velocity, it will experience the same gravitational acceleration as the first stone but with a different force due to its greater mass.
The distance traveled by an object in projectile motion can be calculated using the equation:
s = ut + (1/2)at^2
where:
s = distance traveled or height
u = initial velocity
t = time
a = acceleration
In this case, we want to find the height reached by the second stone. Since the acceleration due to gravity (a) remains constant (-9.8 m/s^2), we only need to compare the initial velocities (u).
For the first stone:
u1 = V
For the second stone:
u2 = 2V
Since the stones are thrown straight up, their final velocities will be zero at their respective peak heights. Therefore, the time taken to reach the peak height will be the same for both stones.
Using the equation of motion, we can calculate the time taken to reach the peak height (t):
0 = u + at
For the first stone:
0 = V - 9.8t1
Solving for t1 gives us:
t1 = V/9.8
For the second stone:
0 = 2V - 9.8t2
Solving for t2 gives us:
t2 = 2V/9.8
Now, let's calculate the height reached by the second stone (H2):
H2 = u2t2 + (1/2)at2^2
Plugging in the values:
H2 = (2V)(2V/9.8) + (1/2)(-9.8)(2V/9.8)^2
Simplifying:
H2 = 4V^2/9.8 - V^2/9.8
Combining like terms:
H2 = 3V^2/9.8
Therefore, the height reached by the second stone (with a mass of 2M and initial velocity of 2V) will be 3V^2/9.8.