PROJECTILE MOTION EXAMPLES

Projectile Motion Examples (1 of 2): Projectile Motion at 30°, 45° and 60° Launch Angles

In the first projectile motion example, a steel ball is going to be launched with a projectile launcher at various launch angles including 30, 45 and 60 degrees. What are the flight distance predictions for each of these launch angles?

Given: Before starting calculation of predicted values, a measurement has been done with the projectile launcher. A steel ball has been launched at 90 degrees (vertically up), maximum height that the ball reaches has been measured as hmax= 3.07 m. The uncertainty of the height measurement is +/- 0.15 m and this is approximately 5% error.

Solution: By using measurement result of 90 degrees launch angle, V02 can be calculated with the following way.

$$For\quad \alpha ={ 90 }^{ \circ }\rightarrow { h }_{ max }=\frac { { { V }_{ 0 } }^{ 2 } }{ 2g }$$ (Eq-1)

$${ h }_{ max }=3.07\pm 0.15\quad m$$  (Measurement Result)

If the measurement result is substituted into Eq-1, we get

$${ V }_{ 0 }^{ 2 }=60.2\pm 3{ { \quad m }^{ 2 } }/{ { s }^{ 2 } }$$.

The flight distance for different launch angles can be predicted by using following formula.

$$Δx=\frac { 2{ { V }_{ 0 } }^{ 2 }\cos { \alpha } \sin { \alpha } }{ g } =\frac { { { V }_{ 0 } }^{ 2 }\sin { 2\alpha } }{ g }$$  (Eq-2)

Since flight distance is proportional to V02, it has exactly same uncertainty value with V02, and consequently h.

The projectile launcher can be set to a launch angle with the accuracy of ± 1deg. In other words, the launch angle α has ±1deg uncertainty. Since flight distance is proportional to Sin2α, it’s needed to calculate Sin2(α± 1) for each launch angle case.

Launch Angle [α]Sin2(α+1)sin2αsin2(α-1) Max. Error (%)
30°Sin(62°) = 0.8829Sin(60°) = 0.8660Sin(58°) = 0.84802
45°Sin (92°) = 0.9994Sin(90°) = 1Sin(88°) = 0.99940.06
60°Sin (122°) = 0.8480Sin(120°) = 0.8660Sin(118°) = 0.88292

For 45° launch angle, angle adjustment error is 0.06% and it’s very small compared to 5% velocity error so it can be neglected. For 30° and 60° launch angles, the angle error is approximately 2 % so it shall be taken into account in the calculations.

The predictions for the range for different launch angles are as follows.

• Range prediction for 30° : Δx= (60.2*0.8660)/(2*9,81) ± 7% = 5.31± 0.37 m
• Range prediction for 45° : Δx= (60.2*1)/(2*9,81)          ± 5% = 6.14± 0.31 m
• Range prediction for 60° : Δx= (60.2*0.8660)/(2*9,81) ± 7% = 5.31± 0.37 m

The projectile motions for 30°, 45° and 60° launch angles are shown below. The initial velocity is same for all the cases. As seen from the animation, the projectile motion with 60° launch angle has the highest and 30° launch angle has the lowest peak height before falling.

The steel ball launched at the 30° angle reached the ground first because its vertical velocity is the lowest (V0Sin30). The steel ball launched at the 60° angle reached the ground last since its vertical velocity is the highest (V0Sin60). As described above that launch cases of 30° and 60° angles travel same distances, the motion with 60° launch angle will have intuitively much longer flight time.

Projectile Motion Examples (2 of 2): Projectile Motion and Free Fall

A man aims his tennis ball at an apple held by a second man who stands on a hill a distance away d as shown below. At the instant the tennis ball is thrown, the second man releases the apple to cause the first man to fail to hit the apple. Show that the second man made the wrong move. The effect of air resistance is ignored.

Solution: The 1st trajectory shown in the figure would be the one without the gravity and  2nd trajectory with gravity. At a certain moment t1, let’s assume that the tennis ball is at A in 1st trajectory. If there is gravity, at the same moment t1 the ball must be at B because xt1 is same for both trajectories since the horizontal velocities are same. The position in x direction is independent whether there is gravity or not.

Let’s now check the position equation in y direction.

$${ y }_{ t }={ (V }_{ 0 }\sin { \alpha } )t-\frac { 1 }{ 2 } g{ t }^{ 2 }$$

If there is no gravity, second term in the above equation doesn’t exist. Therefore this distance between point A and B is 0.5gt12.

Now let’s check the movement of the apple. At time t=0, the apple was released, so in t1 seconds it will have fall down exactly the distance 0.5gt12. So this means the apple and tennis ball will be exactly at same location after t1  seconds and the apple will be hit. If the man had not released the apple, it wouldn’t have been hit.