Question

A ball is thrown up with a speed of 5 m/s. How high will it go before it begins to fall? [h= 1.27 m]

Answer 1

Answer:

$\boxed{\mathfrak{Height \ (h) = 1.27 \ m}}$

Given:

Initial speed (u) = 5 m/s

Final speed (v) = 0 m/s

Acceleration due to gravity (g) = -9.8 m/s²

To Find:

Height (h) reached by ball before it begins to fall

Explanation:

$\sf From \ 3^{rd} \ equation \ of \ motion: \\ \boxed{ \bold{ {v}^{2} = {u}^{2} + 2gh}}$

Substituting values of v, u & g in the equation:

$\sf \implies {0}^{2} = {5}^{2} + 2( - 9.8)h \\ \\ \sf \implies 0 = 25 - 19.6h \\ \\ \sf \implies 19.6h = 25 \\ \\ \sf \implies h = \frac{25}{19.6} \\ \\ \sf \implies h = 1.27 \: m$

$\therefore$

Height (h) reached by ball before it begins to fall = 1.27 m

Answer 2

Answer:

1.25 m or 1.27 m

Explanation:

Given that, a ball is thrown up with a speed of 5 m/s. (u is 5 m/s, at highest point final velocity i.e. v becomes 0 m/s and a is -10 m/s² as it is against the motion)

We have to find the height or distance.

Using the third equation of motion,

v² - u² = 2as

Substitute the values,

→ (0)² - (5)² = 2(-10)s

→ 0 - 25 = -20s

→ 20s = 25

Divide by 20 on both sides,

→ 20s/20 = 25/20

→ s = 1.25

Hence, the height or distance covered by the ball is 1.25 m.

If we take a is 9.8 them s is 1.27 m.