Question

A ball is thrown verticall upward at the Velocity of 30mls. Caluulate the hight covered by the ball in 5 seconds?​

Answer 1

Answer :-

S = 25 m

Given :-

u = 30 m/s

g = 10 m/s²

To find :-

The distance travelled between last second.

Solution:-

Since, ball is thrown upward.

v = 0 m/s.

Time taken to reach the maximum height:-

→t = \dfrac{u}{g}t=

g

u

→t = \dfrac{30}{10}t=

10

30

→t = 3st=3s

Time of Ascend = Time of descent

The distance covered by ball is given by:-

\huge \boxed{ S =ut +\dfrac{1}{2}gt^2}

S=ut+

2

1

gt

2

Distance covered by ball in last 2s:-

→S'= 0 \times 2 +\dfrac{10}{2}(2) ^2S

′

=0×2+

2

10

(2)

2

→S'= 0 + 5 \times 4S

′

=0+5×4

→S'= 20S

′

=20

→S' = 20mS

′

=20m

Distance covered by ball in last 3s :-

→S" = 0 \times 3+\dfrac{10}{2}(3 )^2S"=0×3+

2

10

(3)

2

→S" = 5\times 9S"=5×9

→S"= 45S"=45

→S"= 45mS"=45m

The distance covered by ball is :-

→S = S"- S'S=S"−S

′

→S = 45- 20S=45−20

→S = 25 mS=25m

hence,

Distance covered by ball is 25 m.

Answer 2

the height of ball is 45 metre