Question

A particle is projected with velocity of 10 m/s at an
angle of 15° with horizontal. The horizontal range
will be (g = 10 m/s2)
(1). 10 m
(2) 5 m
(3) 2.5 m
(4) 1 m
​

Answer 1

Answer:

5m

Explanation:

horizontal range = u^2sin2teta/g

u=10m/s

teta=15°

=10^2sin2(15)/10

=100×sin30/10

=100 × 1/2/10

=50/10

=5m

Answer 2

Answer:

(2) 5m

Explanation:

R = u^2 sin 2@/g

R = 10*10*sin(2*15)°/10

R = 100sin 30°/10

R = 100* 1/2/10

R = 5m