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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?

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

Answer :

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

Learn more on time of flight here;

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Rewritten by : Barada

The toy rocket will reach the ground approximately 2.03 seconds after launch.

To find the time it takes for the toy rocket to reach the ground after being launched straight up, we can use the kinematic equation for the position of an object under constant acceleration:
[tex]h(t) = h0 + v0t + 1/2at^2[/tex]
Where:
- h(t) is the height of the rocket at time t,
- h0 is the initial height (3 ft in this case),
- v0 is the initial velocity (60 ft/s in this case),
- a is the acceleration due to gravity ([tex]-16 ft/s^2[/tex] in this case),
- t is the time.
Plugging in the given values:
[tex]h(t) = 3 + 60t - 8t^2[/tex]
When the rocket reaches the ground, the height is 0 ft. So we set h(t) = 0 and solve for t:
[tex]0 = 3 + 60t - 8t^2\\8t^2 - 60t - 3 = 0[/tex]
Using the quadratic formula:
[tex]t = (-(-60)±7 \sqrt{((-60)^2 - 48(-3)))}/(2*8)t = (60 ± \sqrt{(3600 + 96))}/(16)t = (60 ± \sqrt{(3696))}/(16)t = (60 ± 60.8)/(16)[/tex]
This gives two solutions:
t1 ≈ 2.03 s
t2 ≈ -0.78 s
Since time cannot be negative, we take the positive value as the time it takes for the toy rocket to reach the ground after launch. Therefore, the toy rocket will reach the ground approximately 2.03 seconds after launch.