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:

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

Explanation:

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