Question

a body is thrown up with a velocity 29.23ms distance travelled in last second of upward motion is
i) 2.3 m
ii)6 m
iii)9.8m
iv)4.9m

Answer 1

Final Answer: iv) 4.9 m 

Solution: 
Steps:
1)  We will find time to cover upward motion.(i.e last second ).
 We have, 
Initial velocity , u =  29.93 m/s
At max height, final velocity ,v =0
acceleration due to gravity ,a =-9.8 m/$s^{2}$ 

Then ,By Newton' s third equation of motion,
 v= u +at 
=>0= 29.23m/s -9.8*t 
=> t = 2.98s
 
2) Displacement in n'th second , :- Displacement in 'n' seconds -Displacement in (n-1) seconds 
[tex] S_{nth} = S_{n} - S_{n-1} \\ \:\:\:\:\:\:\:\: = [un + \frac{1}{2} a n^{2} ] - [u(n -1 )+ \frac{1}{2} a (n-1)^{2} ] \\ \:\:\:\:\:\:\:\: = u + \frac{1}{2}a(2n-1) [/tex]
     
3) Here, 
u = 29.23m/s 
a = -9.8m/$s^{2}$ 
n = 2.98s 

Therefore, distance covered in 'nth' second :
 [tex]S_{nth} = u + \frac{1}{2}a(2n-1) \\ = 29.23 - \frac{9.8}{2} (2*2.98-1) \\ = 29.23 - 24.30\\ = 4.9 m(approx) [/tex] 

Hence ,Displacement in last second of upward motion  is 

[tex] \boxed{4.9 m} [/tex]

Answer 2

Answer:

4.9 Meters

Explanation:

now the given info

u=29.23 m/s

v=0 m/s

using the 1st eqn of motion V=u+at

0=29.23-gt. .......(because it it projected vertically upward a=-g)