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If a projectile is shot at an angle of #pi/6# and at a velocity of #24 m/s#, when will it reach its maximum height??

Answer 1

Adrian Ayala

#t~~1.223 " seconds"# to 2 decimal places

Note that #9.81# for g is an approximate figure so the solution is an approximation to close limits.

will be 0 at maximum height

#=> (dh)/(dt)~~ 24sin(pi/6)-9.81t#

But at maximum height #(dy)/(dt)=0# giving

#9.81t~~24sin(pi/6)#

#=> t=24/9.81sin(pi/6)#

But #sim(pi/6)=1/2#

#=>t~~12/9.81#

#t~~1.223 " seconds"# to 2 decimal places

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Answer 2

Charlotte Aaron

The projectile will reach its maximum height at the time given by the formula: (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 (approximately (9.8 , \text{m/s}^2)). Substituting the given values: (t_{\text{max}} = \frac{24 \sin(\pi/6)}{9.8}). Calculate to find the time.

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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.