Isaac throws an apple straight up (in the positive direction) from 1.0 m above the ground, reaching a maximum height of 35 meters. Neglecting air resistance, what is the ball’s velocity when it hits the ground?
V^2=Vo^2+2a(Xf-Xo)
=O^2+2(-9.8)(35-0)
Vf^2=-686
Square root 686 which equals
Vf= -26.2 m/s
To find the velocity of the apple when it hits the ground, you can use the concept of conservation of energy. Since there is no air resistance, the total mechanical energy of the apple remains constant throughout its motion.
The mechanical energy of an object can be given by the formula:
E = KE + PE,
where E is the total mechanical energy, KE is the kinetic energy, and PE is the potential energy. At the maximum height of 35 meters, the apple has no kinetic energy and all of its energy is potential energy. Thus, at that point, the equation becomes:
E = 0 + mgh,
where m is the mass of the apple, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
When the apple reaches the ground, its potential energy becomes zero since the height becomes zero. Therefore, the equation becomes:
0 + 0 = 0 + mgh,
Simplifying the equation, we have:
mg(35) = 0,
Finally, we can solve for the velocity by using the equation for free fall:
v = √(2gh),
where v is the velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the maximum height (35 meters). Plugging in the values, we have:
v = √(2 * 9.8 * 35),
v ≈ 25.1 m/s.
Therefore, the velocity of the apple when it hits the ground is approximately 25.1 m/s in the positive direction.
V^2=Vo^2+2a(Xf-Xo)
=O^2+2(-9.8)(35-0)
Vf^2=-686
Square root 686 which equals
Vf= -26.2 m/s
in this equation, why is it 35-0 not 35-1?? if its thrown initially 1m from the ground
h(t) = 1 + vt - 16t^2
max height is reached at t = v/32, so
1 + v(v/32) - 16(v/32)^2 = 35
v = 8√34
h(t) = 1 + 8√34 t - 16t^2
h=0 at t=2.94
v(t) = 8√34 - 32t, so when h=0,
v = 8√34 - 32(2.94) = -47.43 ft/s