Question

question 6 (13 marks) a stone was dropped off a cliff and hit the ground with a speed of 98 m/s (meter/second). what is the height of the cliff? how long does it take for the stone to hit the ground? (hint: the acceleration is -9.8 m/s²)

Explanation:

Step1: Determine the kinematic equation

We use the kinematic equation \( v = v_0 + at \), where \( v \) is the final velocity, \( v_0 \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. The stone is dropped, so \( v_0 = 0 \, \text{m/s} \), \( v = - 98 \, \text{m/s} \) (negative because it's downward), and \( a = - 9.8 \, \text{m/s}^2 \).
Substitute into the equation: \( - 98 = 0 + (- 9.8)t \)

Step2: Solve for time \( t \)

Solve \( - 98 = - 9.8t \) for \( t \). Divide both sides by \( - 9.8 \): \( t=\frac{-98}{-9.8}=10 \, \text{seconds} \)

Step3: Use another kinematic equation for height

Use the equation \( h = v_0t+\frac{1}{2}at^2 \). Substitute \( v_0 = 0 \), \( a = - 9.8 \), and \( t = 10 \):
\( h=0\times10+\frac{1}{2}\times(-9.8)\times(10)^2 \)
\( h = 0 - 4.9\times100=- 490 \, \text{meters} \). The negative sign indicates direction, so the height of the cliff is \( 490 \, \text{meters} \)

Answer:

The time to hit the ground is \( \boldsymbol{10} \) seconds, and the height of the cliff is \( \boldsymbol{490} \) meters.