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A projectile is shot at a velocity of #27 m/s# and an angle of #pi/12 #. What is the projectile's maximum height?

Answer 1

Nathan Ashcroft

The maximum height is #=2.49m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=27*sin(1/12pi)#

Acceleration is #a=-g#

At the maximum height, #v=0#

We apply the equation of motion

#v^2=u^2+2as#

to calculate the greatest height

#0=(27sin(1/12pi)^2-2g*h#

#h=1/(2g)*(27sin(1/12pi))^2#

#=2.49m#

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Answer 2

Parker Abbott

To find the maximum height of the projectile, you can use the following formula: ( H = \frac{{v^2 \sin^2(\theta)}}{{2g}} ), where ( v ) is the initial velocity (27 m/s), ( \theta ) is the launch angle (π/12 radians), and ( g ) is the acceleration due to gravity (9.8 m/s²). Plugging in the values:

( H = \frac{{(27)^2 \times (\sin(\frac{\pi}{12}))^2}}{{2 \times 9.8}} )

( H ≈ 9.72 ) meters

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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.