Question

3.
A ball is projected vertically upward with a speed of 50 ms. Then the speed at half of the maximum
height is (g = 10ms?)
a) 100 ms
b) 125 ms
c) 35 ms
d) 45 ms​

Answer 1

Answer:

(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:

v

2

=u

2

−2gh

max

⇒0=50

2

−2×10×h

max

⇒h

max

=

20

2500

=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:

V

2

=50

2

−2g

2

125

⇒V

2

=2500−1250=1250

⇒V=

1250

=35.35m/s.