Question

A man is on a 30m cliff and throws a ball down with a speed of 7 m/s. How high above the ground is the ball after the ball has fallen for 1 second?

Answer 1

Answer:

210

Explanation:

A ball rolls horizontally off the cliff at a speed of 30 m/s. It takes 7 seconds for the ball to hit the ground. What is the height of the cliff and the horizontal distance traveled by the ball?

S = (1/2)*9.8 m/s^2 * 7^2 = 240.1 m if the ball is very dense so air resistance, and therefore terminal velocity, can be ignored.

S = v * t = 30 m/s * 7 s = 210 m for the horizontal distance, again assuming negligible air resistance.

Answer 2

Given: A man is on a 30 m cliff and throws a ball down with a speed of 7 m/s.

To find: Distance above the ground that the ball has fallen for in 1 second.

Solution:

• Speed of an object is the distance travelled by it in unit time.
• It is given by the formula,

        $s = \frac{d}{t}$

• Here, s is the speed of the ball, d is the distance travelled and t is the time taken.

        $7 ms^{-1} = \frac{d}{1 s}$

        $d = 7 m$

• So, the distance travelled in 1 second after the ball has fallen is 7 m.
• The distance above the ground is calculated as,

        $30 - 7 = 23m$

Therefore, the distance above the ground that the ball has fallen for in 1 second is 23 m.