Question

please solve correctly.​

Answer 1

Answer:

Time to reach maximum height can be obtained from v=u+at

0=20+(−10)t

t=2s

s=ut+0.5at ^2

=20(2)+0.5(−10)(2) ^2

=20m

Thus, total distance for maximum height is 45 m

s=ut+0.5at ^2

45=0+0.5(10)(t )2t =3s

Total time= 3+2= 5s

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Answer 2

Answer:

For upward motion,

Initial velocity (u) = 20m/s

Final veliocity (v) = 0m/s

Accleration due to gravity (g) = 10m/s^2

By applying v - u = at

0 - 20 = 10×t

20 = 10t

t = 2sec

By applying v^2 - u^2 = 2as

400 = 2×10×s

400 = 20s

s = 20m.

For downward motion,

Initial velocity = 0m/s

Accleration due to gravity (g) = 10m/s^2

time = 2+x = 8, x = 8-2 = 6sec.

S = (20+x)m

S = ut+1/2at^2

20+x = 1/2×10×36

20+x = 180

x = 180-20

x = 160m.