Question

A balloon is released from the top of a platform that is 50 meters tall. The balloon rises at the rate of 4 meters per second. write the rule for x in terms of y.

How high is the balloon after 20 seconds?

How high is the balloon after 8 seconds?

Answer 1

y= 50 + 4x, 130 meters and 82 meters.

Explanation:

Given, a balloon is released from a 50 meter tall building and rises at the rate of 4 meters per second. So, the equation will be y = 50 + 4x, where x is time in seconds.

So, after 20 seconds, the balloon will be at

y = 50 + 4x= 50 + 4 X 20 = 50+ 80= 130 meters. Here, x = 20 seconds.

For, x = 8 seconds,

y = 50 + 4x= 50+4 X 8 = 50+32= 82 meters. So the balloon will be at 130 meters and 82 meters respectively.