Question

1500 ms
A particle is projected from a tower as shown in figure, then the distance from
the foot of the tower where it will strike the ground will be:..
loo,
(take g = 10 m/s)
org: 2000
4A74000/3 m
(B) 5000/3 m 2600X
D) 3000 m to
1500m
9oo.
C) 2000 m
77777777777777ITIT​

Answer 1

Answer:

4000/3 m

Explanation:

Particle is projected at 37° with horizontal

Sin37° = 0.6 => Cos37° = 0.8

Then Horizontal Velocity  =  (500/3)Cos37° = 400/3 m/s

Vertical Velocity = (500/3)Sin37° =100 m/s

S  = ut + (1/2)at²

S = Vertical Distance = 1500 m

u = 100 m/s a = g = 10m/s

1500  =  100t  + (1/2)(10)t²

=> 1500 = 100t + 5t²

=> t² + 20t -300 = 0

=>  t² + 30t -10t - 300 = 0

=> (t + 30)(t-10) = 0

=> t = 10

Horizontal Distance = (400/3) * 10 = 4000/3 m

Answer 2

Answer:

x=ut

=500/3×4/5×10

=4000/3

so option a is correct.

value of t = 10 sec comes by second equation of motion that is s=ut+1/2at^2