Question

A stone is thrown in a vertically upward direction with a velocity of 5 m/s² if the acceleration of stone during its motion is 10 m/s² in the downward direction,what will be the highest attained by the stone and how much will it take to reach there?​

Answer 1

♣GIVEN♣

• Initial velocity (u) = 5 m/s

• Acceleration (a) = - 10 m/s² (As stone is coming downward)

• Final velocity (v) = 0 (As stone thrown upwards)

♣TO FIND♣

• Time taken (t)

• Distance covered (s)

♣SOLUTION♣

❶ USING 1st EQUATION OF MOTION :-

• v = u + at

→ 0 = - 5 + (- 10) × t

→ - 5 = - 10 t

→ 5/10 = t

→ t = 1/2

→ t = 0.5 seconds

• Hence, the time taken by stone to reach there is 0.5 seconds.

❷ NOW USING 3rd EQUATION OF MOTION :-

• v² - u² = 2as

→ (0)² - (5)² = 2 × (- 10) × s

→ - 25 = - 20 × s

→ - 25/- 20 = s

→ s = 1.25

• Hence, the height attained by stone is 1.25 seconds.

♣ANSWER♣

• The time taken by the stone to reach there = 0.5 seconds

• The height attained hy the stone = 1.25 metres

Answer 2

Answer:

hope it helps mate

Explanation:

Given, initial velocity, u=5ms  −1

 Final velocity, v=0

Since, u is upward & a is downward, it is a retarded motion.  

∴a=−10ms −2

 Height attained by stone, s=?

Time take to attain height, t=?

(i) Using the relation, v2 −u2=2as, we have

s=v2 −u2 /2a

=(0)^2 −(5)^2 /2×(−10)

=1.25m

(ii) Using the relation, v=u+at

0=5+(−10)t or

t=5/10=0.5s