Question

1. A ball is thrown vertically upward with a speed of 25.0 m/s. (a) How high does it rise? (b) How long does it take to reach its highest point? (c) How long does the ball take to hit the ground after it reaches its highest point? (d) What is its velocity when it returns to the level from which it started? 2. It is possible to shoot an arrow at a speed as high as 100 m/s. (a) If friction is neglected, how high would an arrow launched at this speed rise if shot straight up? (b) How long would the arrow be in the air? 3. Red, blue and yellow ball In my lecture, I gave the two approaches of kinematics for Y-direction: they are 1) free fall (Y YO- 4.9tA2); and 2) Lunch from the ground (Ya Vy0-4.9t 2). They are come from one general formula: 1 Could you use this general formula to derive the shooting down the ball from high board (like 20 meters board diving). In this problem, the Ha-20.0 m; Pyo -5.0m; ay g-9.8( Pyo has same direction as the g). (Red ball) Questions: 1) How many seconds the ball touches down the ground; 2) What is touch down speed? 2. Could you use this general formula to derive the shooting up the ball from high board (like 20 meters board diving). In this problem, the Ho- 20.0 m; "yo -5.0 m; ay 89. has opposite direction as the g). (Blue ball). 3. For a free fall problem, y yo(t-0)-Ho,it is obvious. For lunching from the ground problem, y yelt-O) yground ground)O is not obvious, it is quadratic equation for y0, you can solve groune and t -O, please solve tground, further, for y a, you also can solve two t values, tup and tdon (up is corresponding the ball cross the line of y a when it is going up, down is corresponding to the cross the line of y a again when is going down) (Yellow ball)

Answer 1

Here I am answering first four subparts of question no. 1

It is a case of motion with constant acceleration for a freely falling body. When a ball is thrown vertically upward then at its highest point the velocity will be zero and the ball will start falling with acceleration due gravity g.

Calculation work is shown below.

When ball goes up, acceleration due to gravity 'g' acts downward So a = -g = -9.8 m/s^2 Using equation of motion: v^2 - u^2 = 2as v = final velocity = 0 u = initial velocity = 25 m/s a = -9.8 m/s^2 and S = H Now substitute, 0 - 25^2 m^2/s^2 = 2 times - 9.8 m/s^2 times H H = 625/2 times 9.8 m = 31.88 m = > Ball reaches a height of 31.88 m Time taken to reach the highest Point: use equation S = ut + 1/2 at^2 u = 25 m/s, S = H = 31.88 m, a = -9.8 m/s^2 31.88 = 25 times t - 1/2 times 9.8 times t^2 31.88 = 25 t - 4.9 t^2 = > 4.9 t^2 - 25t + 31.88 = 0 = > t = 2.59 s Time taken to reach ground from highest Point: H = 31.88 m, for downward motion of ball acceleration is positive i.e. a = g = 9.8