If a projectile is shot at a velocity of #2 m/s# and an angle of #pi/4#, how far will the projectile travel before landing?

Answer 1

it will land about 0.638 m far.

The equations of motion need to be recorded in writing:

#x= x_0+vcosthetat# #y= h_0+ v sintheta t + (at^2)/2 #

where #v# is the velocity of the projectile, #theta# is the angle of the trajectory, #h_0# the initial quota of the projectile, #x_0# its initial position.

We possess:

We have #v# and #thetha#: we have to find t from the first equation and substitute its value in the second one. When the projectile lands, it reaches the ground so we have y=0 at that moment. The equations of motion at the landing are:

#x= vcosthetat# #0= v sintheta t - (gt^2)/2 #

from the first equation #t=x/(vcostheta)#. The other becomes: # v sin theta * x/(vcostheta) - g/2*(x/(vcostheta))^2=#

Performing the calculation we eventually have: #x= sqrt( (2v^2 costheta sintheta)/g )~~ 0.638 \ m #

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Answer 2

Layla Ahmed

The projectile will travel approximately 1.02 meters before landing.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.