Question

plz ans fast...need urgent​

Answer 1

Answer:

1.) 1720m

2.) 8.11 m/s^2

3.) 33.124 m

Explanation:

ans 1.- Initial speed of airplane u=0 m/s

Acceleration of the plane a=3.2 m/s^2and time taken t=3.28 s

Using formula, distance covered S=ut+1/2 at^2

⇒ S=0(32.8)+0.5(3.2)(32.8) ^2

=1721.3∼1720m

ans2.-The initial velocity of the Jeep u = 0

Acceleration a = ?

time of journey t = 5.21 sec.

distance covered s = 110 m

s = ut + (1/2)at^2

110 =0+ (1/2) a x5.21 x 5.21

a = (110 x 2)/(5.21 x5.21)

a = 220/27. 14 = 8.11 m/s^2

ans3.- T= 2.6sec

g= 9.8 m/s2

v= u+at

v= 0 + (9.8)(2.6)

v =25.48m/s

s=ut+(1/2)gt2

s= 0 + (1/2) x 9.8 x (2.6)2

=33.124 m

I hope it is helpful for you