Need Help Solving a Physics Problem? T=____?

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SUMMARY

The discussion revolves around solving a physics problem involving the motion of a tennis ball thrown vertically upwards. The initial height is 2 meters, the initial velocity is 9 m/s, and the acceleration due to gravity is modeled as -10 m/s². By applying the kinematic equation \(y = y_0 + v_0 t + 0.5 a t^2\) and solving the resulting quadratic equation \(5t^2 - 9t - 2 = 0\), the time \(T\) when the ball hits the ground is determined. The positive root of the equation, which is approximately 2 seconds, is the valid solution for the time of flight.

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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of motion under gravity, particularly in solving real-world problems involving projectile motion.

cake81
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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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