A plane, diving with constant speed at an angle of 46.0° with the vertical, rele

Question

A plane, diving with constant speed at an angle of 46.0° with the vertical, releases a projectile at an altitude of 699 m The projectile hits the ground 6.09 s after release. (a) What is the speed of the plane? (b) How far does the projectile travel horizontally during its flight? What were the magnitudes of the (c) horizontal and (d) vertical components of its velocity just before striking the ground? (State your answers to (c) and (d) as positive numbers.) UnitsT m/s (a) Numbe122.27 (bNumbe535.64 (c) NumberT89.95 (d) Number 62.59 Units Tm Units T m/s UnitsT m/s

Explanation / Answer

We have following data:

h = 699 m (vertical distance covered after release)

t = 6.09 s (time taken to reach projectile to ground)

Acceleration due to gravity, g = 9.8 m/s2

Now, we can calculate initial speed of projectile in Vertical direction = uv

h = uv * t + (1/2) * g * t2

So, uv = [ h -  (1/2) * g * t2] / t

= (699 - 0.5 * 9.8 * 6.092) / 6.09

uv = 84.94 m/s

As the angle of Plane with vertical was 46o when projectile was released,

Now, we also know the vertical component of Initial velocity of projectile, uv = 84.94 m/s

So, horizontal component of Initial velocity of projectile, uh = uv * Tan (46o)

uh = 84.94 * 1.0355 = 87.955 m/s

Initia; velocity of projectile = Velocity of Plane ; u = uv / cos 46o

So, Speed of Plane; u = 84.94 / 0.695 = 122.27 m/s   (Answer for part a)

Horizontal travel by projectile = uh * t = 87.955 * 6.09 = 535.65 m  (Answer for part b)

As we do not consider any force in Horizontal direction of projectile,

Horizontal speed of projectile when it hits the ground will remain same as it was during release.

Horizontal component of projectile when it hits ground;vh = uh = 87.955 m/s (answer for part c)

Vertical component of projectile when it hits ground; vv = uv + g* t

vv = 84.94 + 9.8 * 6.09 = 144.62 m/s (answer for part d)

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