Question

Stone is thrown vertically upward with the velocity of 25 m/s. How long does it take to reach the maximum distance? Also calculate the distance.​

Answer 1

Answer:

Explanation:

time to Max height = 2.5 s

maxi height = 31.25 m

Explanation:

initial velocity = u = 25 m/s

final velocity = v = 0 (at max height)

acceleration = a = -10 m/sq-sec

a/c eqn of motion,

v = u + at

=> t = (v - u)/a = -25/(-10) = 2.5 sec

again,

v^2 = u^2 + 2as

=> s = (v^2 - u^2)/2a = - 25 × 25 / (-2 × 10) = 31.25 m

max height = 31.25 m

Answer 2

Explanation:

Initial Velocity u=40

Fianl velocity v=0

Height, s=?

By third equation of motion

v2−u2=2gs

0−402=−2×10×s

s=20160

⇒s=80m/s

Toatl distance travelled by stone = upward distance + downwars distance =2×s=160m

Total Diaplacement =0, Since the initial and final point is same.