If a projectile is shot at an angle of #(pi)/8# and at a velocity of #13 m/s#, when will it reach its maximum height?

Answer 1

Nathan Ashcroft

#t=(24.70)/(19.62)=1.26 " "s#

#t=(v_i^2*sin^2 alpha)/(2*g)=(v_y)^2/(2g)#

#v_y=v_i*sin alpha=13*sin (pi/8)=4.97#

#t=(4.97)^2/(2*9.81)#

#t=(24.70)/(19.62)=1.26 " "s#

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Answer 2

Charlotte Aaron

To find the time it takes for the projectile to reach its maximum height, we can use the formula for the vertical component of velocity at any given time:

$v_y = v_0 \sin(\theta) - gt$

At maximum height, the vertical component of velocity is zero, so we set $v_y = 0$ and solve for $t$ :

$0 = 13 \sin(\frac{\pi}{8}) - 9.8t$

$t = \frac{13 \sin(\frac{\pi}{8})}{9.8}$

$t \approx 0.77$ seconds

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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.