Question

If a ball is thrown upwards with a velocity of 40m/s.Then the distance covered by the ball at the end of 6s from the point of projection is (take g=10m/sq)
1)60m
2)0m
3)80m
4)100m​

Answer 1

Given:-

☆Initial velocity of the ball(u)=40m/s

☆Time(t)=6s

☆Acceleration due to gravity(g)=10m/s²

To find:-

☆Distance covered by the ball at the end of 6s.

Solution:-

Using the 1st equation of motion,we get"-

=>v=u+at

=>0=40+(-10)t

=>-40= -10t

=>t= -40/-10

=>t= 4s

Thus,maximum height reached by the ball:-

=>v²-u²=2as

=>0-1600=2×(-10)s

=>s= -1600/-20

=>s=80m

After 4s,the ball will start to move in downward direction,with initial velocity(u)=0

=>s=1/2gt²

=>s=1/2×10×(2)²

=>s=20m

Hence,the total distance covered by the ball is 6s is:-

=>(80+20)m=100m.

Thus,correct option is (4)100m.

Answer 2

Answer:

Given - u = 40 m/s

t= 6 s

v= 0

S = ?

To find - Distance

Solution -

using formula s = ut + 1/2 gt^2

s = 40 × 6 + 1/2 × (-10 )× 6×6

= 240 + 5×36

= 240- 180

= 60 m

Height is (-60 m)