Question

please answer these questions

Answer 1

HELLO THERE!

15. Let's solve this question step by step, applying concept.

First, look at the diagram.

h = + 85 m

g = +

10 m/s²

u (-) = ?

Hence, from:

$S = ut + \frac{1}{2}gt^{2}, \\\\\implies 85 = -u\times5 + 5\times25 \\\\\implies 85 = -5u + 125 \\\\\implies 5u = 40 \\\\\implies u = 8 ms^{-1}$

So, we get the initial velocity with which the ball was thrown.

(i) Now, consider only the first part of the motion, i.e., the ball, starting with u = 8m/s, goes up to a height h, and stops (final velocity = 0).

Then, taking the upward direction as positive and downward direction negative, from:

$v^{2} = u^{2} - 2gh, \\\\\implies 0 = (8)^{2} - 2\times10\times h \\\\\implies 20h = 64 \\\\\implies h = \frac{64}{20} = 3.2m$

Or, the ball goes up by 3.2m from the top of the tower, so it goes up by 85 + 3.2 m = 88.2 m high from the ground (maximum height attained by the ball from the ground is 88.2 metres).

(ii) Now, take only the second part of the motion, in which the ball, from a height of 88.2 m above the ground, has to come down with a free fall.

Take the downward direction as positive.

So, g = +10 m/s², u = 0, h(+) = 88.2 m and v(+) = ?

From relation:

$v^{2} = u^{2} + 2gh, \\\\\implies v^{2} = 0 + 2\times10 \times 88.2 \\\\\implies v^{2} = 1764 \\\\\implies v = 42 m/s$

(iii) To find out the time taken to reach the maximum height, go back to the first half of the motion of the ball, in which the ball starts with a velocity of 8 m/s, and rises to a height of 3.2 m in t seconds, after which its velocity becomes zero.

Take the upward direction as positive and downward direction as negative.

g (-) = 10 m/s, u(+) = 8 m/s, v = 0

From relation:

$v = u - gt, \\\\\implies 0 = 8 - 10t \\\\\implies 10t = 8 \\\\\implies t = 0.8 seconds.$

So, the ball takes 0.8 seconds to reach the maximum height.

16. Let the height of the cliff be h metres, and

Let the time taken to reach the cliff be t.

Therefore, it takes t seconds to cover the distance of 34.3 m.

From the relation:

$S_{t} = u + \frac{a}{2}(2t-1) \\\\\implies 34.3 = 0 + \frac{10}{2}(2t - 1) \\\\\implies 34.3 = 10t - 5 \\\\\implies t = 3.93s (= 4s)$

Now,

Using the second equation of motion,

h = ut + ½ gt²

=> h = 0 + ½ × 10 × 4²

=> h = 80 m

So, the height of the cliff is 80 m.

HOPE MY ANSWER IS SATISFACTORY...

Thanks!