Question

1. the diagram below represents a ball projected horizontally with an initial velocity of 20 m/s eastward off a cliff 100 m high.

how many seconds, to the nearest tenth, does the ball take to reach the ground? $t = \sqrt{\frac{2h}{g}}, g = 9.8\\,\text{m/s}^2$

Explanation:

Step1: Identify given values

We know \( h = 100 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \). The formula for time \( t \) is \( t=\sqrt{\frac{2h}{g}} \).

Step2: Substitute values into formula

Substitute \( h = 100 \) and \( g = 9.8 \) into the formula:
\( t=\sqrt{\frac{2\times100}{9.8}} \)

Step3: Calculate numerator

First, calculate \( 2\times100 = 200 \). So now we have \( t=\sqrt{\frac{200}{9.8}} \)

Step4: Divide inside the square root

Calculate \( \frac{200}{9.8}\approx20.408 \)

Step5: Take the square root

Take the square root of \( 20.408 \): \( \sqrt{20.408}\approx4.5 \) (to the nearest tenth)

Answer:

\( 4.5 \)