If a projectile is shot at an angle of #pi/6# and at a velocity of #32 m/s#, when will it reach its maximum height??

Answer 1

#h_m=13,048 m " maximum height"#

#theta =cancel(pi)/6* 180/cancel(pi)=30^o# #t=w_i*sin theta/g# #t=32*sin 30# #t=32*0,5=16 sec" time to maximum height"# #h_m=(v_i^2*sin^2 30)/(2*g)# #h_m=(32^2* 0,5^2)/(2*9,81)=(1024*0,25)/(19,62)# #h_m=256/(19,62)# #h_m=13,048 m " maximum height"#

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Answer 2

Nora Arzate

The projectile will reach its maximum height when its vertical velocity component becomes zero. Using the formula for vertical velocity in projectile motion, we can find the time it takes for this to occur:

[v_y = v_i \sin(\theta) - gt]

[0 = (32 , \text{m/s}) \sin(\pi/6) - (9.8 , \text{m/s}^2) \cdot t]

Solving for (t):

[t = \frac{(32 , \text{m/s}) \sin(\pi/6)}{9.8 , \text{m/s}^2}]

[t \approx 1.64 , \text{s}]

So, it will reach its maximum height approximately 1.64 seconds after being shot.

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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.