Question

A projectile is fired horizontally from a gun that is 59.0 m above flat ground, emerging from the gun with a speed of 370 m/s. (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

Answer 1

Answer:

Explanation:

Given that,

Height H=59m, initial height yo

Final height on the ground is zero I.e y=0

Since it is fired horizontally,

Then, horizontal velocity is

Vox=370m/s

The initial velocity of the vertical component is zero Voy=0m/s

a. Time the projectile spent in air

Using equation of motion

y—yo = Vot —½gt²

0-59= 0•t - ½×9.81t²

-59 = -4.905t²

t² = -59÷ -4.905

t² = 12.029

t =√12.029

t =3.47seconds

b. Horizontal distance

Using range formula

R=Vox•t

R= 370 ×3.47

R =1283.24m

c. Vertical Component of final velocity

Using education of motion

Vy = Voy - gt

Vy = 0 - 9.81×3.47

Vy = -34.02m/s

Vy = 34.02m/s downward