Projectile motion. ballistic motion free fall. Assumptions. The Earth’s surface is flat (a plane) Gravitational acceleration g = 9.81 m/s 2 There is no air resistance ( Later on during the course, we will return to the subject with more general assumptions ).
( Later on during the course, we will return to the subject with more general assumptions )
Approximate solution without a calculator:
Answers: Flight time is 8 sec.
Highest point 80 m
The bullet experiences 2 different types of motions . Horizontally it is in a uniform motion with a constant speed. Vertically its motion is uniformly accelerating with a = - 9.81 m/s2.
1) Divide the initial velocity vector into horizontal and vertical components:
v0 cosα=400*cos35o =327.7 m/s
2. Calculate the flight time. m/s in 35 deg angle.
Vertical velocity is initially 229.4 m/s. It changes with -9.81 every second. => After about 23 seconds velocity is zero and the bullet is at the ceiling point. After about 46 seconds the bullet hits the ground.
Mathematically: The equation for vertical velocity is:
vy = v0 sinα – g t
Substituting vy = 0 we get the time when the bullet is at highest.
t = v0 sinα / g ( here 23.4 s)
3. Calculate the highest point
Vertical velocity is initially 229.4 m/s. At the ceiling point it is 0. The average velocity during the rise is 229.4/2. The highest y- coordinate is 229.4/2 m/s* 23.4 s = 2684 m = 2. 6 km.
Mathematically: The highest point
ymax = v0 sinα / 2* v0 sinα / g
=> ymax = v02 (sinα)2 / (2g)
3. Calculate the horizontal distance of the flight m/s in 35 deg angle.
Horizontal velocity is initially and all the time 327.7 m/s.
Once we know the flight time 46.8 s , it is easy to calculate the distance x = 327.7 m/s * 46.8 s = 15336 m = 15.3 km
x = v02/g sin(2α)
Observations: The horizontal distance is proportional to the square of the initial velocity.
sin(2α) = 1 when α = 45o.
The shot or throw is longest, when the angle is 45 degrees.
( This is true only when the starting and landing levels are same)
In Olympic games a sportsman throws hammer, which starts its flight at 1.8 m height at the speed of 27.0 m/s. The angle is 47 degrees. Calculate the result:
The flight time:
Problem: Olympic winner of shot put threw 21.32. Assume that the angle was 46 degrees and the ball started at 1.8 m height.
Calculate the initial velocity of the throw.
Result: The initial velocity v0 was 13.91 m/s
and the time = 2.206 s