Question

a ball is projected vertically up with a speed of 50m/s
1)find the maximum height
2)find the time to reach the maximum height ​

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

Explanation:

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