Kindly solve this problem in physics kinetics ex 1 no 4

Kindly solve this problem in physics kinetics ex 1 no 4 3. 4. 5 A stone is thrown upwards with a speed v from the top o velocity 3v. What is the height of the tower ? A motorcydist with unifcwn retardation takes 10 s and 20 s to travel successive quarter kilaneter. How much further he will travel before coming to rest? A stone is thrown vertically upwards with a velocity Of 19.6 m]s. After 2 second, another stone is thrown upwards with a velocity of 9.8 rn/s. When and where will these stones collide?

Dear student,
Let ‘u’ be the initial speed of the motorcycle when it covers the first 250 m in 10 s. Let ‘a’ be the acceleration of the motorcycle. The velocity after this 10 second is,

v = u + at

=> v = u + 10a

Now, using,

S = ut + ½ at²

=> 250 = 10u + ½ a(10²)

=> 250 = 10 u + 50a

=> u + 5a = 25 …………………..(1)

For the next 20 s the, initial velocity is v (= u + 10a).

Using, S = vt/ + ½ at^/2

=> 250 = (u + 10a)20 + ½ a(20²)

=> 250 = 20u + 200a + 200a

=> 250 = 20u + 400a

=> 25 = 2u + 40a

=> 2u + 40a = 25 …………………(2)

(2) - 2×(1) => 2u + 40a – 2u – 10a = 25 – 50

=> 30a = -25

=> a = -25/30 = -5/6 m/s²

(1) => u + 5a = 25

=> u = 25 – 5a = 25 + 25/6 = 175/6 m/s

Let v^/ be the velocity after the 20 s.

v^/ = v + at^/

=> v^/ = (u + 10a) + a(20)

=> v^/ = u + 30a

Let, S^/ be the distance travelled after the 20 s before coming to rest.

0 = (v/)² + 2aS^/

=> 0 = (u + 30a)² +2aS^/

=> 0 = (175/6 – 30×5/6)² - 2×(5/6)S^/

=> 0 = (175/6 - 25)² – (5/3)S^/

=> (5/3)S^/ = (25/6)²

=> (5/3)S^/ = 625/36

=> S^/ = 125/12 m
Regards