If a projectile is shot at an angle of #pi/6# and at a velocity of #18 m/s#, when will it reach its maximum height??

Answer 1

#t~=0,92 " s"#

#g=9,82 " m/s^2#
#v_i=18" "m/s#
#alpha=pi/6#
#sin alpha=0,5#
#t=(v_i*sin alpha)/g#
#t=(18*0,5)/(9,81)#
#t=9/(9,81)#
#t~=0,92 " s"#
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Answer 2
Time of reaching at maximum height #t =(usinalpha)/g=(18*sin(pi/6))/9.8=0.91s#
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Answer 3

To find the time it takes for a projectile to reach its maximum height, you can use the formula: ( t = \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 )). Plug in the values: ( v_0 = 18 , \text{m/s} ) and ( \theta = \pi/6 ). Calculate ( t ).

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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.

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.

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.

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.

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