We use the equation of motion for vertical component,
Here, 
is displacement of bullet, 
is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.
Therefore,
Given, 
Substituting the values, we get time of flight
Answer:
(a) The angle of projection is 63 degree.
(b) The velocity of projection is 24.5 m/s.
Explanation:
Height, h = 1 m
horizontal distance, d = 50 m
time, t = 4.5 s
Let the initial velocity is u and the angle is A.
(a) Horizontal distance = horizontal velocity x time
50 = u cos A x 4.5
u cos A = 11.1 .....(1)
Use second equation of motion in vertical direction
Divide (2) by (1)
tan A = 1.97
A = 63 degree
(b) Substitute the value of A in equation (2)
u x sin 63 = 21.8
u = 24.5 m/s
Answer:
Explanation:
1) The time of flight equation for projectile motion can be used here to find total time in air.
t = 2vsin∅ / g
where v is speed, Ф is launch angle
t = 2×4×sin 60 / 9.8
t = 0.71 seconds
2) Distance where it hit the ground is called as range and has the following standard equation
D = v² sin2Ф/g
D = 4²sin 2×60 / 9.8
D = 1.41m
3) Maximum elevation is maximum time reached
h = v² sin²Ф / 2g
h = 4²sin² 60 / 2*9.8
h = 0.61 m
There is no kinetic energy.
Kinetic energy = 1/2(m)v^2
There is no velocity; therefore, there is no kinetic energy.
