Question
A man throws a tennis ball into the air so that, at the instant when the ball leaves his hand, the ball is 2 m above the ground and is moving vertically upwards with speed 9ms^-1 The motion of the ball is modelled as that of a particle moving freely under gravity and the acceleration due to gravity is modelled as being of constant magnitude 10ms^-2 The ball hits the ground T seconds after leaving the man's hand. Using the model , find the value of T. (4 marks) T= seconds
Answer
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Lionel
Elite · Tutor for 8 years
Answer
To solve the equation mentioned above, we use either factorisation, completing the square, or the Quadratic Formula. Since this equation is already in quadratic equation form - standard is ax^2 + bx + c = 0; a = -5, b = 9, c = 2.Substitute these values into the quadratic formula (-b ± sqrt(b^2 - 4ac))/(2a) to get x1(T) and x2(T).And only write down T is the positive root because Time must be a positive quantity. Answer - T = (positive root) seconds
Explanation
The questions says that the man throws the ball vertically upwards with an initial speed of 9 m/s from an initial height of 2 m. From these details, we assume that the only force acting on the ball is gravity pulling it towards the ground, reducing the speed of the ball. Gravity acts downwards (opposite to the direction of velocity) with an acceleration of 10 m/s^2.To find the total time for the ball to reach the ground from the moment it was thrown in the air, we can use the motion equation: Final position = Initial position + (Initial velocity * Time) - 0.5 * Gravity * Time^2Setting Final position to 0 (as the ball will hit the ground) and then solving for Time T. Thus, we have: 0 = 2 + 9T - 0.5*10*T^2