A blue ball is thrown upward with a velocity of 9 m/s from the top of a high cliff. At the same time, a red ball is dropped from the same spot. The red ball hits the ground exactly... A blue ball is thrown upward with a velocity of 9 m/s from the top of a high cliff. At the same time, a red ball is dropped from the same spot. The red ball hits the ground exactly one second before the blue ball. How high is the cliff? [128.7 m] Read more

Steps to Solve

1. Understand the motion equations

For both balls, we can use the kinematic equations of motion. The equation for displacement is given by:

$$ d = v_i t + \frac{1}{2} a t^2 $$

where

• ( d ) is the displacement,
• ( v_i ) is the initial velocity,
• ( a ) is the acceleration (which is ( -9.8 , \text{m/s}^2 ) for both, since they are falling due to gravity),
• ( t ) is the time.

2. Calculate time for the red ball

Let ( t ) be the time taken by the red ball to hit the ground. The red ball is dropped, so its initial velocity ( v_i = 0 ). Thus, the displacement will be:

$$ d_{\text{red}} = 0 \cdot t + \frac{1}{2}(-9.8)t^2 = -\frac{1}{2}(9.8)t^2 $$

3. Calculate the time for the blue ball

The blue ball is thrown upwards with an initial velocity of ( 9 , \text{m/s} ). It reaches its highest point and then falls down. The time taken for it to reach the ground is ( t + 1 ) (since it hits the ground one second after the red ball). Its displacement is given by:

$$ d_{\text{blue}} = 9(t + 1) - \frac{1}{2} \cdot 9.8(t + 1)^2 $$

Substituting displacement as ( -h ) (height of the cliff):

$$ -h = 9(t + 1) - \frac{1}{2}(9.8)(t + 1)^2 $$

4. Set up the equations

Since both displacements describe the same height of the cliff, we can equate their absolute values:

$$ \frac{1}{2}(9.8)t^2 = 9(t + 1) - \frac{1}{2}(9.8)(t + 1)^2 $$

5. Solve for ( t )

Rearranging the equation yields:

$$ 4.9t^2 + 4.9(t + 1)^2 = 9(t + 1) $$

Expanding and simplifying leads to:

$$ 4.9t^2 + 4.9(t^2 + 2t + 1) = 9t + 9 $$

Combining terms:

$$ 9.8t^2 + 9.8t + 4.9 - 9t - 9 = 0 $$

Rearranging gives:

$$ 9.8t^2 + 0.8t - 4.1 = 0 $$

6. Use the quadratic formula

Applying the quadratic formula ( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ):

Here, ( a = 9.8, b = 0.8, c = -4.1 ).

Calculating the discriminant:

$$ b^2 - 4ac = 0.8^2 - 4 \cdot 9.8 \cdot (-4.1) $$

Now, compute the solutions for ( t ).

7. Calculate height ( h )

Once ( t ) is found, substitute it back into either height equation:

$$ h = \frac{1}{2} \cdot 9.8t^2 $$

This will give the height of the cliff.