Question

A projectile is fired at an angle 60degrees to horizontal with an initial velocity of 800metre per second find a,time of flight.b, horizontal range.c, Time of flight to reach maximum height​

Answer 1

Answer:

(a) Time of flight = 80√3 s

(b) distance travelled by projectile = 320√3 m

(c) distance travelled by projectile = 320√3 m

Explanation:

given that,

A projectile is fixed at an angle of 60° to the horizontal with an initial velocity of 800 m/s

here,

initial velocity of the projectile = 800 m/s

angle at which it was thrown = 60°

gravitational acceleration = 10 m/s²

(a) time of flight = 2usin@/g

where,

u = initial velocity

g = gravitational acceleration

putting the values,

time of flight = 2 × 800 × sin60/10

= 1600 × √3/2/10

= 800√3/10

= 80√3 s

so,  

Time of flight = 80√3 s

(b) distance it travels before it hits the ground

here,

to find the range(R)  

so,

R = u² sin2/g

= 800 × 800 × sin120°/10

= 6400 × √3/2 /10

= 320√3 m

so

distance travelled by projectile = 320√3 m

(c) the time of flight for the projectile to reach its maximum height.

here,.  

time if flight for maximum height

= 800 × sin60

= 800 × √3/2

= 400√3

so,.

time of flight for maximum height

= 400√3 m