Question

g
60.
A projectile of mass 500 g is projected with the
speed 50 m/s from the height of 50 m from the
ground at an angle of 60° with vertical. If constant
horizontal opposite force 5 N is acting on the
projectile then time after which the projectile,
returns to the same horizontal level is

(1) 2.5 s
(2)5/√2s

(3) 5s
(4)5√2s

​

Answer 1

Explanation:

(2ucosa/g)

=[(2×50×cos60°)]/g

=5sec

Answer 2

Answer:

Ans: 5sec.

Explanation:

Always remember that Time of flight is affected by vertical component of Force/ Acceleration $\frac{t \: = \: 2u \cos( \alpha )}{g} = \: \frac{2 \times 50 \times \frac{1}{2} }{10} = 5sec$