Question

An object is projected upward with initial velocity 50 m / s. Calculate time period after which it will reach the ground also calculate the height attained by the object. take g=10 m /(s.s). Draw displacement - time graph, velocity- time graph and acceleration- time graph of this motion.


(ANSWER IN DETAIL WITH GRAPH)​

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.