A ball is thrown vertically upward with a velocity of 49 m/s. calculate:(a) The maximum height to which it rises.(b) The total time it takes to return to the surface of the earth. (Take g=9.8m/s2)

Sure, let's solve this step by step:

(a) The maximum height to which it rises:

We can use the following equation of motion to solve this:

v^2 = u^2 - 2gh

where: v = final velocity (which is 0 at the maximum height) u = initial velocity = 49 m/s g = acceleration due to gravity = 9.8 m/s^2 h = height

Substituting the values, we get:

0 = (49)^2 - 29.8h => h = (49)^2 / (2*9.8) => h = 122.5 m

So, the maximum height to which the ball rises is 122.5 m.

(b) The total time it takes to return to the surface of the earth:

We can use the following equation of motion to solve this:

v = u - gt

where: v = final velocity = 0 (when the ball reaches the maximum height) u = initial velocity = 49 m/s g = acceleration due to gravity = 9.8 m/s^2 t = time

Substituting the values, we get:

0 = 49 - 9.8*t => t = 49 / 9.8 => t = 5 s

But this is the time taken to reach the maximum height. The total time taken to return to the surface of the earth is twice this value (as the time taken to ascend and descend under gravity is the same).

So, the total time taken = 2t = 25 = 10 s.

So, the ball takes 10 seconds to return to the surface of the earth.