Question

a cricket ball is thrown vertically upward with an intial velocity of 4ms-1 find .a.Its velocity after three second .b. maximum height attained and the time taken to reach it​

Answer 1

a) - 25.43 m/s

b) H = 0.815 m

t = 0.41 s

Step-by-step explanation:

Given that:

Initial velocity (u) = 4 m/s, and the acceleration due to gravity (g) = 9.81 m/s²

a) For a body thrown upward, gravity acts on the body and causes a decrease in velocity as it continues to move, making the body to reach a maximum height and start falling. The formula for a vertically thrown object is:

v = u - gt

t = time = 3 s and v = final velocity.

Therefore:

v = 4 - (9.81 × 3) = - 25.43 m/s

b) The equation is given as:

v² = u² - 2gH

Where H is the maximum height. At maximum height, final velocity v = 0. Therefore:

2gH = u²

H = u² / 2g

H = 4² / 2(9.81) = 0.815 m

Given that v = u - gt and at maximum height, v = 0

Therefore t = u / g = 4 / 9.81 = 0.41 s