if a penny is dropped from the top of a 320 ft. building, how fast will it be moving when it hits the ground?
v2 = u2 + 2gH,
Here, u = initial velocity = 0
v = final velocity
g = 32ft/s2
H= 320ft
So, v2= 0 + 2 × 32 × 320
or,. v = ✓(2×32×32×2×5)
or, v = 64✓5 ft/s
is this correct?
looks good
Yes, your calculation is correct. The final velocity of the penny when it hits the ground will be 64√5 ft/s.
Yes, your calculation is correct. To find the final velocity of the penny when it hits the ground, you used the equation v^2 = u^2 + 2gH, where u is the initial velocity (which is 0 because the penny is dropped), g is the acceleration due to gravity (32 ft/s^2), and H is the height of the building (320 ft).
Plugging in the given values, you correctly substituted u^2 = 0, g = 32 ft/s^2, and H = 320 ft into the equation.
Simplifying the equation gives:
v^2 = 0 + 2 * 32 * 320
v^2 = 64 * 320
Taking the square root of both sides of the equation gives:
v = √(64 * 320)
Evaluating this expression gives:
v ≈ √(20480) ≈ 64√5 ft/s
So, the final velocity of the penny when it hits the ground is approximately 64√5 ft/s.