A toy rocket is launched straight up into the air with an initial velocity of 60ft/s from a table 3 ft above the ground. If acceleration due to gravity is -16ft/s^2, approximately how many seconds after the launch will the toy rocket reach the ground?

To solve this problem, we just have to an equation of motion and proceed to find the time of flight. The time of flight of the projectile is 0.05s

Time of Flight

The time of flight of this projectile can be calculated using an equation of motion.

Data;

Initial velocity = 60ft/s
Height = 3ft
Acceleration due to gravity = -16ft/s^2

The negative acceleration shows that the projectile is moving against gravity at that time of it journey.

Using the equation

[tex]s = ut+\frac{1}{2}at^2\\[/tex]

Let's substitute the values and solve

[tex]s = ut+\frac{1}{2}at^2\\\\3 = 60*t + \frac{1}{2}(16)t^2\\3 = 60t +8t^2\\8t^2 + 60t -3 = 0[/tex]

Solving the quadratic equation, the values are -7.45s and 0.05s and in the given options, the time of flight of the projectile is 0.05s

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