Question

What is the maximum height reached by a ball, if it is projected with an initial velocity of 4i + 3j?
Give me explanation ​

Answer 1

Answer:

Explanation: H = u^2(in y direction)/2g

g=10

initial velocity in

y direction is 3

so H = 9/2x10

= 0.45

+

Answer 2

The maximum height reached by a ball is 0.45 meters.

Explanation:

Given that,

Initial velocity of the ball, $u=(4i+3j)\ m/s$

At maximum height, final velocity of the ball we be 0, v = 0

Let h be the maximum height reached by the ball. It can be calculated using third equation of motion as :

$v^2-u^2=2ah$

Here, a = -g

$u^2=2gh$

$h=\dfrac{u_y^2}{2g}$

Since, $u_y=3\ m/s$

$h=\dfrac{(3\ m/s)^2}{2\times 9.8\ m/s^2}$

h = 0.45 meters

So, the maximum height reached by a ball is 0.45 meters. Hence, this is the required solution.

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