need help with this

Question

A blue ball is thrown upward with a velocity of 9 m/s upward from the top of a high cliff. At the same time, a red ball is dropped from the same spot. The red ball is observed to hit the ground below exactly one second before the blue ball. How high is the cliff? 

Answer 1

The blue ball will to up to a maximum height of 9²/2g= 4.13 m above the launch height at time = 9/g= .917 seconds. The time to reach the ground that height =sqrt(2*(hcliff+4.13)/g). Now, the time for the red ball to reach the ground = sqrt(2*hcliff/g). Note that the difference between these two times = one second. This yields an equation sqrt(2*hcliff/g)-sqrt(2*(hcliff+4.13)/g)=1 which can be solved for hcliff = 16.2 meters

Answer 2

red ball. blue ball

V_i = 0. V_i = 9m/s

a = g a = g. g = -9.8m/s²

Displacements equal; time for red ball = t, time for blue ball = t+1

since displacements equal write the equation for both balls y = V_it + .5gt² and set equal to each other and solve. For red ball y= 0 + .5(-9.8) t² → y= -4.9t² ; for blue ball y= 9(t+1) .5(-9.8)(t +1)²→ y= 9t + 9 - 4.9(t+1)²

Since y same for both balls -4.9t² = 9t + 9 -4.9[t² + 2t + 1]

finish the algebra and solve for t then substitute it into displacement equation of red ball, y = -4.9t²

displacement is negative because ball falls downward from cliff, which has a positive height.

hope this helps please give me some feedback.

Answer 3

There are many ways to approach this problem. My first take is to find the critical time by saying the final y value of both balls must occur at the appropriate times:

y₀ - 1/2 g(t-1)² = y₀ + v₀t - 1/2 gt²

solve for t (note the y₀ and 1/2 gt² terms cancel)

Once you have t, you can solve for y0 by using either equation.

Good luck!