Question

A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to fall? (g = 9.8 m/s2)

Answer 1

Answer:

approx. 11.48m

Explanation:

u=15 g=-9.8

v=0at highest point

v^2-u^2=2gs

Answer 2

$\huge\star\sf\underline\pink{Solution :-}$

We have,

$\sf{Initial\:speed\:of\:ball (u) = 15\:m/s } \\ \\ \sf{Final\:speed\:of\:ball (v) = 0} \\ \\ \sf{Acceleration\:due\:to\: gravity (g) = -9.8\:m/s^2 \: (Retardation)}$

$\sf\underline\red{To\:Find:-} \\ \\ \: \: \: \: \bullet\sf{Height (h) = ?}$

Now,

Putting Values into the Formula ,

$\sf\underline\red{Formula:-} \\ \\ \\ \blue{\boxed{\sf{v^2 = u^2 + 2gh}}} \\ \\ \\ \longrightarrow\sf{ (0)^2 = (15)^2 + 2 \times (-9.8) \times h} \\ \\ \\ \longrightarrow\sf{ 0 = 225 - 19.6 \times h} \\ \\ \\ \longrightarrow\sf{ 19.6h = 225} \\ \\ \\ \longrightarrow{\sf{ h = \dfrac{225}{19.6} }} \\ \\ \\ \longrightarrow\sf{ h = 11.4 \:m}$

Thus,

The ball will go to a maximum Height of 11.4 metres.