Question

A ball is thrown vertically upwards with a velocity of 50 m/s. calculate
(a) the maximum height to which it rises.
(b) the total time it takes to return to the surface of earth. (takeg =10 m/s2 )

Answer 1

Given :

Initial velocity of ball = 50m/s

To Find :

• Maximum height attained by ball.
• Total time taken by ball to return to the surface of the earth.

Solution :

■ For a body thrown vertically upward, g is taken negative as it acts in the opposite direction.

Acceleration due to gravity has constant magnitude so we can apply equation of kinematics to solve this question$.$

A] Max. height attained :

❇ Third equation of kinematics :

➠ v² - u² = 2gH

• At maximum height; v = 0

➠ 0² - 50² = 2(-10)H

➠ -2500 = -20H

➠ H = 2500/20

➠ H = 125 m

B] Time of flight :

❇ First equation of motion :

Time of flight for a body thrown vertically upward at a speed of u is given by, T = 2u/g

➛ T = 2u/g

➛ T = 2(50)/10

➛ T = 100/10

➛ T = 10 s

Answer 2

Answer:

hope it helps

Explanation: