A ball is thrown vertically upward froma height of 40 m and hits the groundwith a speed that is three times itsinitial speed. What is the time taken (insec) for the fall (Take 𝑔 = 10 𝑚/𝑠2) ?

To solve this problem, we need to use the equations of motion.

Step 1: Identify the given values Initial height (h) = 40 m Acceleration due to gravity (g) = 10 m/s² Final velocity (v) = 3u (where u is the initial velocity)

Step 2: Use the first equation of motion to find the initial velocity (u) The first equation of motion is v = u + gt. Since we are considering the upward motion of the ball, g will be negative (-g) as it is acting against the motion of the ball. When the ball reaches its maximum height, its final velocity will be 0. So, we can set v = 0 and solve for u:

0 = u - gt u = gt

Step 3: Use the second equation of motion to find the time of flight (t) The second equation of motion is h = ut + 0.5gt². Substituting the value of u from step 2, we get:

h = gtt + 0.5gt² 40 = 10t² + 0.510*t² 40 = 15t² t² = 40/15 t = sqrt(40/15) = 2.58 seconds

Step 4: Use the third equation of motion to find the time of fall The third equation of motion is v² = u² + 2gh. Since the ball hits the ground with a speed that is three times its initial speed, we can set v = 3u and solve for h:

(3u)² = u² + 21040 9u² = u² + 800 8u² = 800 u² = 100 u = 10 m/s

Now, we can use the first equation of motion to find the time of fall:

v = u + gt 3u = u + 10t 2u = 10t t = 2u/10 = 2*10/10 = 2 seconds

So, the time taken for the fall is 2 seconds.