# Answer to Question #12278 in Atomic and Nuclear Physics for krantikiran

Question #12278

a body is thrown vertically upwards with initial velocity of 100m/s find (1)time taken to reach at maximum height (2)maximum height attend (3)velocity after 5 second (4)velocity after 16 second. Take (a=10m/s2)

Expert's answer

(1) time taken to reach at maximum height

tm = V0 / g = 100 / 10 = 10

s

(2) maximum height attend

h = V0 * t - g * tm^2 / 2 = 100 * 10 - 10 *

10^2 / 2 = 1000 - 1000 / 2 = 500 m

(3) velocity after 5 second

t = 5 <

tm = 10, so:

V(5) = V0 - g * t = 100 - 10 * 5 = 50 m/s

(4) velocity after

16 second

t = 16 > tm = 10, so:

V(16) = 0 + g * (t - tm) = 0 + 10 * (16

- 10) = 60 m/s

tm = V0 / g = 100 / 10 = 10

s

(2) maximum height attend

h = V0 * t - g * tm^2 / 2 = 100 * 10 - 10 *

10^2 / 2 = 1000 - 1000 / 2 = 500 m

(3) velocity after 5 second

t = 5 <

tm = 10, so:

V(5) = V0 - g * t = 100 - 10 * 5 = 50 m/s

(4) velocity after

16 second

t = 16 > tm = 10, so:

V(16) = 0 + g * (t - tm) = 0 + 10 * (16

- 10) = 60 m/s

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