A projectile is shot from the ground at a velocity of #1 m/s# and at an angle of #(pi)/2#. How long will it take for the projectile to land?

Answer 1

#v_f = v_i + at#
#pi/2# is straight up
#0 = 1 - (9.81 xxt) # and #t= 1/9.81 ~~ .1 s#

graph{-(50x)^2 +5 [-1, 1, 0, 5]} Observe the arbitrary graph that I provided. It show that the projectile will climb up and come down after it reach maximum height. At maximum height the speed #v_f = 0# knowing this write #v_f = v_i + at# we know that #v_i = 1m/s; and a = g = -9.81 m/s^2# Thus #0 = 1 - (9.81 xxt) # and #t= 1/9.81 ~~ .1 s#

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Answer 2

Elena Albarran

To find the time it takes for the projectile to land, we can use the equation for the horizontal motion of a projectile: ( t = \frac{2v_0\sin(\theta)}{g} ), where ( v_0 ) is the initial velocity, ( \theta ) is the angle of projection, and ( g ) is the acceleration due to gravity (approximately 9.8 m/s²). Plugging in the given values, we get: ( t = \frac{2 \times 1 \times \sin(\pi/2)}{9.8} = \frac{2}{9.8} ) seconds. Solving this equation gives us ( t = 0.204 ) seconds. Therefore, it will take approximately 0.204 seconds for the projectile to land.

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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.