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

Answer:

D/H =15

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:

       $H = \frac{v_{0} ^{2} }{2*g} (2)$

• 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:

       $D = \frac{15*v_{0} ^{2} }{2*g}$

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

       $H = \frac{v_{0} ^{2} }{2*g}$

• ⇒ D = 15 * H

        $\frac{D}{H} = 15$