Question

A ball is thrown vertically upwards with a velocity of 30 m/s. If the acceleration due to gravity is
10 m/s, what will be the distance travelled by it in the last second of motion before again come
to his hand :
(1) 5 m
(2) 10 m
(3) 25 m
(4) 30 m​

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:-

→$t = \dfrac{u}{g}$

→$t = \dfrac{30}{10}$

→$t = 3s$

• Time of Ascend = Time of descent

The distance covered by ball is given by:-

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

Distance covered by ball in last 2s:-

→$S'= 0 \times 2 +\dfrac{10}{2}(2) ^2$

→$S'= 0 + 5 \times 4$

→$S'= 20$

→$S' = 20m$

Distance covered by ball in last 3s :-

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

→$S" = 5\times 9$

→$S"= 45$

→$S"= 45m$

The distance covered by ball is :-

→$S = S"- S'$

→$S = 45- 20$

→$S = 25 m$

hence,

Distance covered by ball is 25 m.

Answer 2

Explanation:

u=30m/s^2(given)

g=10m/s^2 (given)

=》t=u/g

=》t=3u/10

=》t=3 sec

FORMULA USED:-

{ S=ut+1/2gt^2}

=》S'=0×2+10/2 (2^2)

=》S'=20 m

DISTANCE COVERED BY LAST 3 SEC.

=》S"=0×3+10/2 (3^2)

=》S"=5×9

=》S"=45m

DISTANCE COVERED BY BALL IS:-

S=S"-S'

S=45-20 m

S=25 m