Question

A ball is thrown vertically upwards with a
velocity of 30 m/s. If the acceleration due to
gravity is 10 m/s2, what will be the distance
travelled by it in the last second of motion.
​

Answer 1

Distance  traveled by it in the last second of motion is 25 m

Given

A ball is thrown vertically upwards with a  velocity of 30 m/s

Acceleration due to  gravity is 10 m/s²

To Find

distance  traveled by it in the last second of motion

Concept Used

As the acceleration due to gravity is constant throughout the motion , so we need to apply equation's of motion ,

⇒ v = u + at

⇒ s = ut + ¹/₂ at²

⇒ v² - u² = 2as

⇒ Sₙ = u + ᵃ/₂ ( 2n-1 )

Solution

Given ,

• Initial velocity , u = 30 m/s
• Acceleration due to gravity , a = - 10 m/s²

[ ∵ thrown against gravity ]

• Final velocity , v = 0 m/s

[ ∵ stops finally ]

Apply 1st equation of motion ,

⇒ v = u + at

⇒ (0) = (30) + (-10)t

⇒ 10t = 30

⇒ t = 3 s

Now , Distance moved by the particle in nth second is given by ,

$\bf \pink{\bigstar\ \; S_n=u+\dfrac{a}{2}(2n-1)}$

• u = 0 m/s
• n = 3
• a = 10 m/s²

$\to \rm S_3=(0)+\dfrac{10}{2}(2(3)-1)\\\\\to \rm S_3=5(5)\\\\ \to \bf S_3=25\ m$

Distance  traveled by it in the last second of motion is 25 m