A field team is trying to remotely place a sensor in the caldera of a semi-activ

Question

A field team is trying to remotely place a sensor in the caldera of a semi-active volcano, the caldera being too hot and noxious for people to venture onto. Their launch site is on the rim, 43 m above the level floor of the caldera.
The first sensor was launched at a speed of 23 m/s, 60.0° above the horizontal. According to the last readings taken before it was destroyed, it hit the surface of the caldera at 37 m/s, which was apparently too much for it to take. Also, it landed 65 m from the rim, and a location closer to the center (farther from the rim) would be preferable. the second sensor, the team decides to keep the launch speed at 23 m/s, but lower the launch angle to 45°. Their reasoning is that, since the second projectile won't rise as high as the first, its impact speed will be slower, allowing it to survive. Additionally, since the launch angle is closer to horizontal, the sensor will travel farther horizontally before impact, thus getting closer to the center of the caldera.

Question: will the plan for the second launch work, or does it need to be revised? Prove your conclusion by calculation.

Explanation / Answer

First of all we assume that there is no air resistance and the value of g=9.8 and then we start calculations.
Conservation of energy :-
.5mv12=.5mv22+mgh

.5v12=.5*232+9.8*43

so v1= 37m/s

so our assumption was right that there wasn't any air resistance.

then it's pointless to throw it with the same speed becaue the impact speed will not depend upon the angle but it willdepend upon initial speed and height.

now time of flight :-

-43 = 23sin(60)*t -.5*9.8*t2

so, t= 5.625sec

now d=23cos(60)*5.625 = 64.6875m

for the second case

time of flight :-

-43 = 23sin(45)*t -.5*9.8*t2

t= 5.054sec

now d=23cos(45)*5.054 = 82.2m

But the secont statement is not right if we continue increasing angle the horizontal distance will shorten because time of flight will be less.

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