This Free Fall and Projectile Motion Physics question asks how to solve a projectile problem when a body is launched at 24.5 m/s at 30 degrees below the horizontal, and why the missing launch height changes the answer for time of flight, maximum height and range.
Question from Cybeir Kim Otchia: I cannot find the answer to the following question: A body is projected with an initial velocity of 24.5 m/s at 30 degrees below the horizontal. Find (a) time of flight, (b) maximum height attained by the body, and (c) range. I would appreciate your help with this. Thank you in anticipation.
Answered by John
Short Answer
The question is incomplete as written. Because the body is projected 30 degrees below the horizontal, it is already moving downwards from the start. To find the time of flight and range, we need to know the height of the launch point above the landing point.
If the body is launched and lands at the same height, the only valid contact time is immediate: time of flight is 0 seconds, range is 0 metres, and the maximum height above the launch point is 0 metres.
If the old answer of about 2.5 seconds and 53 metres was intended, then the missing assumption is that the body was launched from about 61.25 metres above the ground. Without that height, the problem does not have one unique numerical answer.
Step 1: Split The Initial Velocity Into Components
Projectile motion becomes easier when the velocity is split into horizontal and vertical components. The initial speed is 24.5 m/s, and the angle is 30 degrees below the horizontal.
The horizontal component is:
vx = 24.5 cos 30 degrees = 21.2 m/s
The vertical component is downward, so if upward is positive:
vy = -24.5 sin 30 degrees = -12.25 m/s
This sign matters. A positive launch angle would send the body upwards first. This question gives a downward launch angle, so the body starts falling immediately.
Step 2: Understand The Maximum Height
For a normal projectile fired upwards, the maximum height is found when the vertical velocity becomes zero. That is not what happens here.
Because the initial vertical velocity is already downward, the body never rises above the launch point. Therefore, the maximum height above the launch point is:
Maximum height above launch point = 0 metres
If the question asks for maximum height above the ground, then the answer is simply the original launch height. But that height was not supplied in the question.
Step 3: Write The Vertical Motion Equation
Let h be the height of the launch point above the landing point. With upward taken as positive, the vertical position is:
y = h – 12.25t – 4.9t2
The body lands when y = 0, so:
0 = h – 12.25t – 4.9t2
Rearranged:
4.9t2 + 12.25t – h = 0
This is why the missing height matters. Without h, the equation cannot produce one fixed time of flight.
Step 4: Use The General Formula
Solving the quadratic equation gives the positive time of flight:
t = [-12.25 + sqrt(12.252 + 19.6h)] / 9.8
Once the time is known, the horizontal range is:
Range = 21.2t
That is the clean way to solve the problem. The height must be supplied first, then the time and range follow.
Why 2.5 Seconds Only Works With A Missing Height
A time of flight of 2.5 seconds only makes sense if the body was launched from about 61.25 metres above the landing point. That height is not stated in the original question, so it has to be treated as an extra assumption.
Using h = 61.25 m:
0 = 61.25 – 12.25t – 4.9t2
At t = 2.5 seconds:
61.25 – 12.25(2.5) – 4.9(2.5)2 = 0
So, with that added height assumption, the time of flight is 2.5 seconds.
The horizontal range is then:
Range = 21.2 x 2.5 = about 53.0 metres
The maximum height is still 0 metres above the launch point, because the body is projected downwards from the start. If height is measured from the ground instead, the maximum height would simply be the launch height: 61.25 metres.
Why The Original Question Is A Trap
The wording looks like a normal projectile motion question, but it leaves out the height. That missing detail changes the answer completely.
If the projectile is launched from a cliff, window, platform or aircraft, the time of flight depends on that starting height. If it is launched from ground level at 30 degrees below the horizontal and lands on the same ground level, it has no meaningful flight path before hitting the ground.
So the proper answer is not just a number. The proper answer is to state the missing assumption, then solve from there.
Final Answer
As written, the question cannot be answered uniquely because the launch height is missing.
If the body starts and lands at the same height, the time of flight is 0 seconds, maximum height above launch point is 0 metres, and range is 0 metres.
If the intended hidden assumption is that the body starts about 61.25 metres above the landing point, then the time of flight is 2.5 seconds, the range is about 53.0 metres, and the maximum height is 0 metres above the launch point.
Disclaimer: This is educational physics guidance, not a substitute for a teacher’s marking scheme. For coursework, always state assumptions clearly, show your working, and follow the method expected by your course or examiner.
Sources And Further Reading
- OpenStax – Projectile motion equations and components.
- Khan Academy – Two-dimensional projectile motion explained.
- The Physics Classroom – Horizontal and vertical velocity components.
- Physics LibreTexts – Projectile motion and gravity.
Join The Discussion
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