Question

a ball is thrown with a velocity 30 ms-1 from the ground. how high it will rise. calculate the time taken to reach the ground.​

Answer 1

v²-u²=2gh

0-(30)²=-2*10*h

-900=-20h

h=45m

h=ut-1/2g(ta)²

45=30t-5(ta)²

(ta)²-6(ta)+9=0

(ta)²-3(ta)-3(ta)+9=0

((ta)²-3)²=0

ta=3s

h=1/2g(td)²

45=5(td)²

(td)²=9

td=3s

t=(ta)+(td)

t=3s+3s

t=6s

Answer 2

Given,

Initial velocity(u) = 30m/s

To Find,

Height up to which the ball rise and time taken by the ball to reach the ground.

Solution,

Using the equation of motion,

v²-u² = 2gh

0-30² = 2(-10)(h)

900 = 20h

h = 45 m

Now, the time taken to reach the top.

v = u+at

0 = 30-10(t)

t = 3 s.

Time taken to reach at ground = 2*3 = 6 s

Hence, the height up to which the ball rise is 45 m and the time taken to reach the ground is 6 sec.