Question

A projectile was launched horizontally with a velocity of 468m/s, 1.86m above the ground how long did it take the projectile to hit the ground

Answer 1

Answer:

0.616 s to 3 significant figures

Explanation:

Using the equation s=ut+0.5at^2 in the vertical direction:

1.86=0×t+0.5gt^2

1.86=4.905t^2

t^2=1.86÷4.905

t=0.6157961457

=0.616 s (3sf)

Answer 2

We have that for the Question "A projectile was launched horizontally with a velocity of 468m/s, 1.86m above the ground how long did it take the projectile to hit the ground" it can be said that the time to heat the ground is

t=0.616s

From the question we are told

A projectile was launched horizontally with a velocity of 468m/s, 1.86m above the ground how long did it take the projectile to hit the ground

Generally the equation for the vertical distance   is mathematically given as

d=0.5*9.8

d=4.9

Therefore

Time

$t=\sqrt{\frac{1.86}{4.9}}$

t=0.616s

Hence the time to heat the ground is

t=0.616s

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