Question

a ball is thrown vertically upwards from the top of a cliff of 21.21m with an initial velocity 11m/s. Taking g to be 9.8 m/s^2 .i) calculate the height to which it will reach before heading to the ground .ii) the velocity with which it will reach the ground. iii) the total time of the journey​

Answer 1

Answer:

Pleasemark Brainlist

Explanation:

29.4

Answer 2

Answer: i)6.173m ii)23.156m/s iii)3.48s

Explanation:



maximum height means v=0, taking g as negative when ball is going upwards

v²-u²= -2gh

-11²=. -2*9.8*h

i)121/19.6= hmax=6.173m

total height before reaching ground= 21.21 + 6.17 = 27.38m, Case 2

 g is positive since now ball is moving downwards

u i.e initial velocity = 0 ,since it is at hmax

v²-u²=.2gh

v²=.√2*g*h

ii)v= final velocity= √2*9.8 *2727.38 = 23.156m/



total time= time to reach h max + time to reach the

time to reach hmax

g will be negative since going upwards

=> v=u-gt

0= 11-9.8×t.

11/9.8 = t

 time to reach ground

g will be positive since motion is downwards

u will be 0 since we are returning from hmax

=> v=u+gt

 23.156= 9.8*t

t= 23.156/ 9.8

iii)total time=(2.3.156+ 11)/9.8= 3.48s