A stone is thrown in a vertically upwad direction with a velocity of 5m/s . If the acceleration of the stare duing it's motion is 10m/s2 in the down word direction. What will be the height attained by the stone and how much tione will it take to reach there
Answers
Given that, a stone is thrown in a vertically upwad direction with a velocity of 5m/s means the initial velocity of the stone is 5 m/s. The acceleration of the stone duing it's motion is 10m/s² in the downword direction.
At the highest point, the final velocity of the stone is 0 m/s.
We have to find the height attained by the stone and time taken by the stone to reach there.
From above data we have; u = 5m/s, v = 0 m/s and a = -10 m/s² (as it is against the gravity)
Using the Third Equation Of Motion,
v² - u² = 2as
Substitute the known values,
→ (0)² - (5)² = 2(-10)(s)
→ -25 = -20s
→ 5 = 4s
→ 1.25 = s
Therefore, the distance covered by the stone is 1.25 m.
Now, using the First Equation Of Motion,
v = u + at
Substitute the values,
→ 0 = 5 + (-10)t
→ 5 = 10t
→ 0.5 = t
Therefore, the time taken by the stone is 0.5 sec.
Answer :
Given that a stone is thrown vertically upward direction with a velocity of 5m/s. Acceleration during its motion is 10 m/s² in the downward direction.
We have to find height attained by stone and how much time taken by stone to reach there.
From the data we have :
Initial speed (u) = 5 m/s
Acceleration (a) = -10 m/s² [As it's thrown upwards ]
Final speed (v) = 0 m/s [At the highest point velocity/speed will be zero]
Now using 1st equation of motion to find time taken to reach there :
→ v = u + gt
→ 0 = 5 + (-10)t
→ 0 = 5 - 10t
→ 10t = 5
→ t = 5/10
→ t = 0.5 s
∴ Time taken to reach highest point = 0.5 seconds.
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Now using 2nd equation of motion to find height attained by stone :
→ h = ut + ½ gt²
→ h = 5 × 0.5 + ½(-10) × (0.5)²
→ h = 2.5 - 5 × 0.25
→ h = 2.5 - 1.25
→ h = 1.25 m
∴ Height attained by stone = 1.25 m