If a projectile is shot at an angle of #(5pi)/12# and at a velocity of #6 m/s#, when will it reach its maximum height??

Answer 1

#t_e~=0,59 s#

#"you can find the elapsed time to maximum height by: "# #t_e=v_i*sin alpha /g# #v_i:"initial velocity of object"# #alpha:"angle of projectile :" alpha=(5pi)/12*180/pi=75^o# #g:"acceleration of gravity"# #t_e=(6*sin 75)/(9,81)# #t_e~=0,59 s#

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Answer 2

William Audy

To find the time it takes for a projectile to reach its maximum height, you can use the formula: ( t_{max} = \frac{v_i \sin(\theta)}{g} ), where ( v_i ) is the initial velocity, ( \theta ) is the launch angle, and ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )). Plugging in the given values, you can calculate the time it takes for the projectile to reach its maximum height.

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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.