Question

a train is travelling at a speed of 72 km/h .on applying brakes ​

Answer 1

Question:

A ball thrown vertically upwards reaches the highest point of its path after 2s. Find the initial velocity with which the ball is thrown. Also, find the height of this highest point. (Accelration due to gravity = 10 m/s²)

Answer:

$\boxed{\mathfrak{Initial \ velocity \ (u) = 20 \ m/s}}$

$\boxed{\mathfrak{Height \ of \ highest \ point \ reached \ by \ ball \ (h) = 20 \ m}}$

Given:

Final velocity (v) = 0 m/s

Acceleration due to gravity (g) = -10 m/s²

Time taken (t) = 2s

To Find:

$\sf \star$ Initial velocity (u)

$\sf \star$ Height of highest point reached by ball (h)

Explanation:

$\sf \star \: From \ 1^{st} \ equation \ of \ motion: \\ \boxed{ \bold{v = u + gt}}$

Substituting values of v, a & t in the formula, we get:

$\sf \implies 0 = u + (- 10)(2) \\ \\ \sf \implies 0 = u - 20 \\ \\ \sf \implies u = 20 \: m/s$

$\therefore$

Initial velocity (u) = 20 m/s

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

Substituting values of v, u & g in the formula, we get:

$\sf \implies {0}^{2} = {20}^{2} + 2( - 10)h \\ \\ \sf \implies 0 = 400 - 20h \\ \\ \sf \implies 20h = 400 \\ \\ \sf \implies h = \frac{400}{20} \\ \\ \sf \implies h = 20 \: m$

$\therefore$

Height of highest point reached by ball (h) = 20 m

Answer 2

Explanation:

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