Assume that the projectile behaves like a particle moving in a vertical coordinate plane, and the only force acting on the projectile during its flight is the constant force of gravity pointing straight down.

Assume that our projectile is launched from the origin at time t = 0 into the first quadrant with initial velocity v₀. (See figure 1)

If v₀ makes an angle a with the horizontal, then v₀ = (v₀ cos a )i + (v₀ sin a )j.

The projectile's initial position is r = 0i + 0j = 0.

Newton's 2^nd Law of Motion says that the force acting on the projectile is its mass m times its acceleration or

r = 0 at t = 0

a = -g j

figure 1

if r is the projectile's position vector and t is time.

If force is solely the gravitational force, mg j, then

Performing the required integrations and substitutions, we determine that

where a is the launch angle and v₀ is the initial speed.

FACT:

x (t) = (v₀ cos a )t and y (t) = (v₀ sin a )t - (g /2)t ²

HEIGHT:

The projectile reaches its highest point when its vertical velocity component is zero.

For this value of t, the value of y is the following.

FLIGHT TIME:

To find when the projectile lands, when fired over horizontal ground, set y = 0 and solve for t.

RANGE:

FACT:

The range is the largest when sin 2a = 1 or a = 45 ^o.

EXAMPLE 1:

A baseball is thrown from the stands, 32 ft above the field at an angle 30^o up from the horizontal. When and how far away will the ball strike the ground if its initial speed is 32 ft/sec? (see figure 2)

SOLUTION:

Gravity for this problem is 32.

figure 2

EXAMPLE 2:

In Moscow in 1987, Natalya Lisouskaya set a women's world record by putting an 8-lb 13-oz shot 73 ft 10 in. Assuming that she launched the shot at a 40 ^o angle to the horizontal 6.5 ft above the ground, what was the shots initial speed. (see figure 3)

SOLUTION:

73 ft 10 in » 73.83 ft

g = 32

x = v₀ cos 40 ^o t

y = 6.5 + (v₀ sin 40 ^o)t - 16t ²

I will solve for t by setting x = 73.83.

figure 3

Plug this value into y. The position of the shot put at that value of t will be zero.

EXAMPLE 3:

What two angles of elevation will enable a projectile to reach a target 16 km down range on the same level as the gun if the projectile's initial speed is 400m/sec? (See figure 4)

SOLUTION:

g = 9.8

16 km = 16000 m

x = (400 cos a ) t

y = (400 sin a ) t - 4.9t ²

I will set x = 16000 and solve for t.

figure 4

When t is the above expression, y will be zero. I will solve the resulting equation for a .

16000tan a - 7840sec ² a = 0

16000tan a - 7840(tan ² a + 1) = 0

-7840tan ² a + 16000tan a - 7840 = 0