Question

A ball is projected vertically upward with a speed of 40 m/s. Find (a) the maximum height, (b) the time to reach the maximum height, (c) the speed at one fourth the maximum height. Take g = 10m/s2

Answer 1

Step-by-step explanation:

(a) Initial velocity of ball (u)=50m/s, acceleration of ball =−g,

final velocity at the highest point (v)=0

So applying the 3rd equation of motion we get:

v2=u2−2ghmax

⇒0=502−2×10×hmax

⇒hmax=202500=125m

(b) Let the time required to reach max height be t. Then applying 1st equation of motion we get:

v=u−gt

⇒0=50−10t

⇒t=5s

(c) Let speed at half of max height be V then:

V2=502−2g2125

⇒V2=2500−1250=1250

⇒V=1250=35.35m/s.