Question

A ball is thrown vertically upwards from the ground with an initial velocity of 50m/s. What is the total time by the ball to get its maximum height? (take g=10)

A. 500s
B. 50s
C. 5s
D. 0.5s​

Answer 1

c

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:

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.

Answer 2

Step-by-step explanation:

initial velocity (u) = 50m/s

final velocity (v) = 0m/s

i.e the ball stops for a while at its maximum height.

g = 10m/s^2

from the 1st equation of vertical upward motion:

v = u - gt

hence t = (v + u)/t

t = (0 + 50)/10

t = 50/10 = 5s.

C. 5s.