Question

A ball thrown down from a balcony lands in 0.7 s on the ground at a speed of 23 m/s. Find:

a) the initial speed;


b) the height from which it was thrown;


c) the time to land if it were thrown up from the balcony with the same initial speed.

Answer 1

Question #55092, Physics / Mechanics | Kinematics | Dynamics |

A ball thrown down from a balcony lands in 0.7 s on the ground at a speed of 23 m/s. Find:

a) the initial speed;

b) the height from which it was thrown;

c) the time to land if it were thrown up from the balcony with the same initial speed.

Answer:

The speed of the ball on the ground is defined by the equation:

$$v = v_0 + gt$$

, where $t$ – the time of fly and $v_0$ – the initial speed.

Thus,

a)

$$v_0 = v - gt = 23 \, \text{m} \, \text{s}^{-1} - [9.8 \, \text{m} \, \text{s}^{-2} \times 0.7 \, \text{s}] = 16.14 \, \text{m/s}$$

b) The height equals:

$$h = v_0 t + 0.5 g t^2 = 16.14 \, \text{m} \, \text{s}^{-1} \times 0.7 \, \text{s} + 0.5 \, [9.8 \, \text{m} \, \text{s}^{-2} \times 0.49 \, \text{s}^2] = 11.298 \, \text{m} + 2.401 \, \text{m} = 13.7 \, \text{m}$$

c) If the ball goes up it travels unless its speed becomes zero:

$$v = v_0 + gt_1 = 0$$

, $t_1$ – the time of fly in the opposite direction to the ground.

Therefore,

$$t_1 = \frac{v_0}{g} = \frac{23}{9.8} \, \text{s} = 2.35 \, \text{s}$$

The ball travels the following distance:

$$h_1 = v_0 t - 0.5 g t_1^2 = 23 \, \text{m} \, \text{s}^{-1} \times 2.35 \, \text{s} - 0.5 \, [9.8 \, \text{m} \, \text{s}^{-2} \times 5.5225 \, \text{s}^2] = 27 \, \text{m}$$

After this the ball falls down with zero initial speed being of $h$ (27 m + 13.7 m) over the ground.

The time of this free fall is found using the equation:

$$h = 0.5 g t_2^2$$

.

Thus,

$$t_2 = \left(\frac{h}{0.5g}\right)^{0.5} = \left(\frac{40.7}{0.5 \times 9.8}\right)^{0.5} = 2.88 \, \text{s}$$

Finally, the total time of fly is:

$$t = t_1 + t_2 = 2.35 \, \text{s} + 2.88 \, \text{s} = 5.23 \, \text{s}$$

http://www.AssignmentExpert.com/