A toy rocket is launched from a platform 2.1 meters above the ground in such a way this it’s height , h (in meters) after t seconds is given by the equation h=-4.9t^2+14.0t + 2.1. How long will it take for the rocket to hit the ground?

Answer 1

To find the time it takes for the rocket to hit the ground, we need to find the value of t when the height, h, is equal to zero.

The equation is: h = -4.9t^2 + 14.0t + 2.1

Setting h to zero: 0 = -4.9t^2 + 14.0t + 2.1

This equation is a quadratic equation, and we can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are: x = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = -4.9, b = 14.0, and c = 2.1.

t = (-14.0 ± √(14.0^2 - 4(-4.9)(2.1))) / (2(-4.9))

Simplifying this equation, we get:

t = (-14.0 ± √(196.0 + 40.92)) / (-9.8)

t = (-14.0 ± √(236.92)) / (-9.8)

Now, we have to consider both the positive and negative solutions since time cannot be negative:

t = (-14.0 + √(236.92)) / (-9.8) or t = (-14.0 - √(236.92)) / (-9.8)

Simplifying these equations, we get:

t ≈ 4.13 seconds or t ≈ 0.312 seconds

Therefore, it will take approximately 4.13 seconds for the rocket to hit the ground.