Question

A projectile is fired at an angle of 60° to the horizontal with an initial velocity of 800 m s⁻¹.a. Find the time of flight of the projectile before it hits the ground.b. Find the distance it travels before it hits the ground (range).c. Find the time of flight for the projectile to reach its maximum height.

Answer 1

Hii dear,

# Answers-
T = 1384 s
R = 27712 m
Ta = 692 s

## Explaination-
# Given-
u = 800m/s
θ = 60°

# Solution-
a) Time of flight-
T = 2usinθ/g
T = 2×800×sin60
T = 1384 s

b) Horizontal range,
R = u^2sin2θ/2g
R = 800^2 × sin120 / (2×10)
R = 32000×0.866
R = 27712 m
R = 27 km

c) Time of ascent-
Ta = usinθ/g
Ta = 800×sin60
Ta = 692 s

Hope that is useful...

Answer 2

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