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A projectile is shot from the ground at a velocity of #15 m/s# at an angle of #(5pi)/12#. How long will it take for the projectile to land?

Answer 1

William Audy

The time is #=2.96s#

Resolving in the vertical direction #uarr^+#

initial velocity is #u=vsintheta=15*sin(5/12pi)#

distance travelled #=0#

Acceleration is #a=-g#

We apply the equation of motion

#s=ut+1/2at^2#

#0=15sin(5/12pi)*t-1/2*g*t^2#

Therefore,

#t=0# which is the initial condition and

#t=2*15/g*sin(5/12pi)#

#=2.96s#

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Answer 2

Adrian Ayala

To determine the time of flight, use the projectile motion formula: ( \text{time} = \frac{2 \times \text{initial velocity} \times \sin(\text{launch angle})}{\text{acceleration due to gravity}} )

Substitute the values: ( \text{time} = \frac{2 \times 15 \times \sin\left(\frac{5\pi}{12}\right)}{9.8} )

Calculate to find the time of flight.

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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.