If a projectile is shot at an angle of #(2pi)/3# and at a velocity of #5 m/s#, when will it reach its maximum height?

Answer 1

time taken to reach maximum height = #0.4 s# (approx)

The velocity vector of the projectile has a x- component and y - component . since we are dealing with height , we just need the y-component . velocity in #y# direction = #5 sin ((2pi)/3) ms^-1# =#(5sqrt 3) /2 ms^-1#
now using initial velocity #u = (5 sqrt3 )/2 ms^-1# ,
final velocity #v = 0 ms^-1#, acceleration due to gravity #a = -9.8 ms^-2# .
we have #v = u + at#
#=> 0 = (5sqrt3) /2 + -9.8 t#
#=> 9.8 t = (5sqrt3) /2#
#=> t = (5sqrt3) /(2×9.8) #
#=> t= 0.4 s # (approx)
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Answer 2

The projectile will reach its maximum height at half of the total flight time. The time to reach maximum height (t_max) can be calculated using 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.

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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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