# PhysicsMechanics help

#### cake81

##### New member
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 modelled 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 dont know how togo about solving this because I dont understand the part with the particle moving freely

#### Euge

##### MHB Global Moderator
Staff member
Welcome, cake81 ! 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?

#### cake81

##### New member
Welcome, cake81 ! 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?

#### Euge

##### MHB Global Moderator
Staff member
Yes. Only the positive answer makes sense.