A watermelon cannon fires a watermelon vertically up into the air at a velocity of +9.5m/s m/s, starting from an initial position 1.2 meters above the ground. When the watermelon reaches the peak of its flight, what is:
(a) its velocity
(b) its acceleration
(c) the elapsed
(d) the height above the ground

Question

A watermelon cannon fires a watermelon vertically up into the air at a velocity of +9.5m/s m/s, starting from an initial position 1.2 meters above the ground. When the watermelon reaches the peak of its flight, what is:
(a) its velocity
(b) its acceleration
(c) the elapsed
(d) the height above the ground

anonymous · Answer

a) Here , at the top of the flight , 

the velocity is zero

anonymous · Answer

b)
as the acceleration of the watermelon is same at all time in the air,

acceleration  = -g = -9.8 m.s^2

anonymous · Answer

c) time taken to reach top is t 

Using first equation of motion 

v =  u + a*t 

0 = -9.5 +  9.8 * t 

t = 0.97 s 

the time taken to reach the top is 0.97 s

anonymous · Answer

d) height above the ground  = v^2/2g + Ho

 height above the ground  =  9.5^2/*2 * 9.8 + 1.2

 height above the ground  = 5.81 m 

the  height above the ground  is 5.81 m

ranya99 · Answer

Thanks a Lot