MHB Need Help Solving a Physics Problem? T=____?

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To solve the physics problem, the initial height of the tennis ball is 2 meters, with an initial upward velocity of 9 m/s and a downward acceleration of 10 m/s². The kinematic equation used is y = y₀ + v₀t + 0.5at², leading to the equation 5t² - 9t - 2 = 0. The solution yields two values for T, but only the positive value is relevant for the time until the ball hits the ground. The final answer for T is confirmed to be 2 seconds.
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Hello! I need help with this question:
A tennis ball is thrown in the air by a man so that at the instant when the ball leaves his hand, the ball is 2m above the ground and is moving vertically upwards with speed9m/s^-1
The motion of the ball is modlledas that of a particle moving freely under gravityand the acceleration due to gravity is modeled as being of constant magnitude of 10 m/s^-1
The ball hits theground T seconds after leaving the hand
Using the model, find the value of T

I don't know how togo about solving this because I don't understand the part with the particle moving freely
 
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Welcome, cake81! (Wave)

We are given an initial height of $y_0 = 2\,\text{m}$, an initial velocity of $v_0 = 9\, \text{m/s}$ in the upwards vertical direction, and an acceleration of $a = -10\,\text{m/s}$ (the acceleration vector faces downwards). Using the kinematic equation
$$y = y_0 + v_0 t + .5 at^2$$ with $y = 0$ we solve $0 = 2 + 9t + (.5)(10)t^2$, or $$5t^2 - 9t - 2 = 0$$ Can you take it from here?
 
Euge said:
Welcome, cake81! (Wave)

We are given an initial height of $y_0 = 2\,\text{m}$, an initial velocity of $v_0 = 9\, \text{m/s}$ in the upwards vertical direction, and an acceleration of $a = -10\,\text{m/s}$ (the acceleration vector faces downwards). Using the kinematic equation
$$y = y_0 + v_0 t + .5 at^2$$ with $y = 0$ we solve $0 = 2 + 9t + (.5)(10)t^2$, or $$5t^2 - 9t - 2 = 0$$ Can you take it from here?

thank you so much. i got 2, i only have to use the positive answer right?
 
Yes. Only the positive answer makes sense.
 
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