A lead ball is dropped into a lake from a cliff 100m from thesurface of the lake

Question

A lead ball is dropped into a lake from a cliff 100m from thesurface of the lake. The ball moves in the water with the samespeed as it had when it struck the surface of the water. It takes 8s to reach the bottom of the lake. How deep is the lake? Please show steps and formulas used to gain answer....thankyou A lead ball is dropped into a lake from a cliff 100m from thesurface of the lake. The ball moves in the water with the samespeed as it had when it struck the surface of the water. It takes 8s to reach the bottom of the lake. How deep is the lake? Please show steps and formulas used to gain answer....thankyou

Explanation / Answer

Neglecting air drag and the Coriolis Force (due to rotation ofEarth), We use kinematics (in y direction) ----- Y = yo + vyot +(1/2)ayt2 0 = 100m + (0m/s)t +(1/2)(-9.81m/s2)t2 t2 = (100m)/(4.91m/s2) t2 = 20.4 s2 t = 4.52 s ----- Ay = ay Vy = ayt + vyo Vy = ayt + 0 Vy = (-9.81m/s2)(4.52s) Vy = -44.3 m/s ------- (indeed, accounting for air drag, the lead ball would likelybe close if not exceeding terminal velocity at 44.3 m/s, which weknow is impossible for an object to exceed terminal velocity whenin free-fall. Also note, the negative denotes direction:downward). -------- Assuming the lead ball moves in water at same speed as it hadwhen it struck surface (-44.3 m/s), which, again, is absurd,because the bouyoncy force of the water is in fact about 1,000times stronger in magnitude than air drag (because water is 1,000times denser than air, and F=ma, such that the given mass of waterper unit volume is 1,000 times greater than that of air for a givenvolume, hence the force would be 1,000 times greater), we nowcalculate the depth of the lake: ------- Kinematics again (with new initial conditions) Y = yo + vyot +(1/2)ayt2 Here, there are no forces acting on the lead ball (an absurdclaim, but one we'll have to live with).
So ay = 0 also yo is the original position of the lake, whichwe'll set to be our origin (y = 0) calculating the depth of the lake, Y Y = 0 + (-44.3m/s)t + 0 Y = (-44.3m/s)(8.00s) Y = -354m ------ Hence a displacement of -354 meters from the surface, countingup to be positive y and down to be negative y, which meansthat the depth of the lake is: ----- Depth of lake = 354 m ----- I hope this solution was a Lightsaber for you. Please remember to always be a Physics Jedi. Here, there are no forces acting on the lead ball (an absurdclaim, but one we'll have to live with).
So ay = 0 also yo is the original position of the lake, whichwe'll set to be our origin (y = 0) calculating the depth of the lake, Y Y = 0 + (-44.3m/s)t + 0 Y = (-44.3m/s)(8.00s) Y = -354m ------ Hence a displacement of -354 meters from the surface, countingup to be positive y and down to be negative y, which meansthat the depth of the lake is: ----- Depth of lake = 354 m ----- I hope this solution was a Lightsaber for you. Please remember to always be a Physics Jedi.

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