Question

a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its launch, then falls back downward and misses the edge of the cliff, landing below, at the bottom of the cliff, with a speed of 4V0. What is the ratio of the cliff height D to the peak height H

Answer 1

D/H =15

Step-by-step explanation:

• We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
• At this point, we can find the value of H, applying the following kinematic equation:

`v_(f) ^(2) -v_(0) ^(2) = 2* g* H (1)`

• If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

`H = (v_(0) ^(2) )/(2*g) (2)`

• We can use the same equation, to find the value of D, as follows:

`v_(f) ^(2) -v_(1) ^(2) = 2* g* D (3)`

• In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
• When we replace these values in (1), we find that v₁ = -v₀.
• Replacing in (3), we have:

`(4*v_(0))^(2) - (-v_(0)) ^(2) = 2* g* D\\ \\ 15*v_(0)^(2) = 2*g*D`

• Solving for D:

`D = (15*v_(0) ^(2) )/(2*g)`

• From (2) we know that H can be expressed as follows:

`H = (v_(0) ^(2) )/(2*g)`

• ⇒ D = 15 * H

`(D)/(H) = 15`