If a projectile is shot vertically from the ground at #1 m/s#, how long will it remain in the air?

Answer 1

#~~0.2s#

Since the projectile is projected vertically then free vertical motion under gravity to be applied. Here Initial velocity #u = 1m/s# If its time of flight be taken as T s, then after Ts it will return to the initial position at the ground ,making a verticalb displacement #h=0# Taking acceleration due to gravity #g =9.8ms^-2# we can write: #h=uxxT-1/2xxgxxT^2=>0=1xxT-1/2xx9.8xxT^2# #:. T=1/4.9s~~0.2s#
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Answer 2

The time the projectile will remain in the air can be calculated using the formula: ( t = \frac{2v}{g} ), where ( v ) is the initial velocity (1 m/s) and ( g ) is the acceleration due to gravity (approximately 9.8 m/s²). Plug in the values 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.

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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