Answer to Question #163404 in Physics for kyra

Question

1. All is thrown vertically upward from a point 5m above ground level with a speed of 15ms^-1. Find the height above this point reached by the ball and calculate the speed with which the ball hits the ground.
2. A particle moves in a straight line with a constant velocity of 12𝑚𝑠^-1 for 3 seconds. It then decelerates at 4 ms^-2 for a period of 6 seconds. It then accelerates to rest in a further time of 3 seconds. Draw the velocity-time graph for the motion of the particle and calculate the total distance travelled during the motion.
3. A particle is projected from horizontal ground with a velocity of 20 m/s at an angle of 60^o to the horizontal. Find the time taken for the particle to hit the horizontal ground, the greatest height reached above the ground and the horizontal distance covered when it hits the ground at the same horizontal level.

Answer 1

1. The height above this 5m-point:

$$h=\frac{v^2}{2g}=11.5\text{ m}.$$

The speed with which the ball hits the ground:

$$v=\sqrt{2(h+y)g}=18.0\text{ m/s}.$$

2. The total distance travelled during the motion:

$$d=v(t_1+t_2)-\frac{a_2t_2^2}{2}+(v-a_2t_2))$$