Question

a ball is rolled of along the edge of a horizontal table with a velocity 4m/s it hits the ground after 0.4 s. then height of table is
​

Answer 1

$\mathcal{\green{\underline{\blue{GIVEN:}}}}$

• Initial velocity (u) = 0ms⁻¹.
• Final velocity (v) = 4ms⁻¹.
• Time (t) = 0.4 s.
• Acceleration due to gravity (a) = 10ms⁻².

$\mathcal{\green{\underline{\blue{TO \:\: FIND:}}}}$

• We need to find the height (s).

$\mathcal{\green{\underline{\blue{SOLUTION:}}}}$

Let's recall the three equations of motion:

$\to \sf{\red{v=u+at}}$

$\to \sf{\red{s=ut+\frac{1}{2}at^2 }}$

$\to \sf{\red{2as=v^2-u^2}}$

It is apt to use the third formula because we are given time, initial velocity, and acceleration, and we have to find the height.

$\to \sf{\greed{2as=v^2-u^2}}$

Substitute the values.

$\sf \to 2 \times 10 \times s= 4^2-0^2$

$\to \sf s=\frac{20}{16}$

$\boxed{\sf{\blue{s=0.8 m.}}}$

HEIGHT OF THE TABLE IS 0.8 METERS.

$\huge\star\:\:{\orange{\underline{\pink{\mathbf{HOPE \:\: THIS \:\: HELPS \:\: :D}}}}}$