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If a projectile is shot at an angle of #pi/12# and at a velocity of #1 m/s#, when will it reach its maximum height??

Answer 1

William Audy

The maximum height is #=0.013m#

The highest point is determined by

#h_(max)=(v_0^2sin^2theta)/(2g)#

#v_0=1ms^(-1)#

#theta=pi/12#

#g=10ms^(-2)#

Thus,

#h_(max)=1*sin^2(pi/12)/(2*10)#

#=0.013m#

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Answer 2

Axel Acevedo

The projectile will reach its maximum height at the time when its vertical velocity component becomes zero. Using the equations of motion, the time to reach maximum height ( t_{\text{max}} ) is given by ( t_{\text{max}} = \frac{V_0 \sin(\theta)}{g} ), where ( V_0 ) is the initial velocity, ( \theta ) is the launch angle, and ( g ) is the acceleration due to gravity. Substituting ( V_0 = 1 ) m/s, ( \theta = \frac{\pi}{12} ), and ( g = 9.8 ) m/s(^2), we can calculate ( t_{\text{max}} ).

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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.