Question

A ball is thrown vertically upward with a velocity 25meterspersecond on its weight down it was caught 5 meters above the ground find how fast it was travelling wh3n kt was caught and also find total time of the trip ?

Answer 1

Question 32301

Let the time for moving upwards until full stop be $t_s$ . If at the moment of stop velocity is zero, then $v = v_0 - gt_s = 0 \Rightarrow t_s = \frac{v_0}{g}$ . The maximum height is $h = v_0 t_s - \frac{gt_s^2}{2} = \frac{v_0^2}{2g}$ . Moving down (from the point of stop – maximum height point) is with no initial velocity, hence the law of motion is $y(t) = h - \frac{gt^2}{2}$ , so the time to move down is

$t_2 = \sqrt{2\frac{h}{g}}$ . Plugging $h = \frac{v_0^2}{2g}$ into latter formula gives $t_2 = \sqrt{\frac{v_0^2}{g^2}} = \frac{v_0}{g}$ . Hence, total time of movement is $t = t_s + t_2 = 2\frac{v_0}{g}$ .

Now, let us find the time needed to move from maximum height point to point of 5 meters above the ground. Distance to travel is $h - 5$ , hence time is

$t = \sqrt{2\frac{(h - 5)}{g}} = \sqrt{2\frac{\left(\frac{v_0^2}{2g} - 5\right)}{g}}$ . The velocity at that moment of time is

$v = gt = \sqrt{2g(\frac{v_0^2}{2g} - 5)} = 22.9\frac{m}{s}$