At what angle should a body be projected with a velocity 24m/s just to pass over the obstacle 14m high at a distance of 24m.
Answers
This is a problem in physics relating to projectile motion. While not exactly what the problem is, we can use a formula for calculating the angle θ to hit coordinate (x,y) as a good start.
Please refer to the attached image for the full formula as it is quite long and complex to write in text.
This formula computes the angle θ of launch to hit a target at range x and altitude y when fired from (0,0) and with initial speed v.
Also, the g in the formula refers to the gravitational acceleration constant = 9.8m/s^2.
The coordinate (0,0) simply refers to the start position.
Plug in the values for each of the variables:
x = 24 (the distance from start of launch to the obstacle in meters)
y = 14 (the height of the obstacle in meters)
v = 24 (the initial speed of the projectile in m/s)
g = 9.8 (gravity constant)
The computation can be quite long.
Note how the formula contains "±" or plus-minus symbol. This means that there are two possible answers. Depending on which one you use, you will get two different angles.
If there aren't any computation errors, you will get the angles 44.6 degrees and 75.7 degrees (rounded to the nearest tenths).
This means that you can launch the projectile at those angles to HIT the top of the obstacle. This is what I meant when I said the formula is not quite the final solution. In order for the projectile to pass OVER the obstacle, we need to know it's size and shape. For example, if it's a ball with a radius of 1m, then we just simply adjust our values such that y = 14m + 1m or y = 15m. Redo the computation using the adjusted value for y and you will get the new angles. When launched at those angles, our imaginary ball's skin will ever so slightly touch the top obstacle and pass over it.
This computation DOES NOT yet take into account air resistance and friction. However, since the problem did not provide information on those, it's safe to ignore them just for learning's sake.
Answer:
1,19/5.
Explanation:
tanθ=19/5
tanθ=1
tanθ=3
tanθ=2
Answer :
A::B
Solution :
x=24=ucosθ.t
⇒t=2424cosθ=1cosθ
y=14=usinθt−12gt2
⇒14=usinθcosθ−5cos2θ⇒14=utanθ−5sec2θ
⇒5tan2θ−24tanθ+19=0⇒tanθ−5sec2θ
⇒5tan2θ−24tanθ+19=0⇒tanθ=1,19/5.
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