Answer:

Introduction:
Projectile motion is a type of motion experienced by objects that are thrown into the air or launched at some angle. The motion of a projectile can be described as a combination of horizontal motion with a constant velocity and vertical motion with a constant acceleration due to gravity. In this problem, we have two projectiles that are thrown from the same point simultaneously with the same velocity of 10 m/s.

Vertical Projectile:
One of the projectiles goes straight vertically. The vertical motion of this projectile can be described by the following equation:

y = uyt - 1/2gt^2

where y is the vertical displacement, uy is the initial vertical velocity, t is the time, and g is the acceleration due to gravity. In this case, the initial vertical velocity is 10 m/s, so we have:

y = 10t - 1/2gt^2

After 1 second, the vertical displacement of this projectile is:

y = 10(1) - 1/2(9.8)(1)^2 = 5.1 m

Projectile at 60 degrees:
The other projectile is launched at an angle of 60 degrees with the vertical. The horizontal and vertical components of the velocity can be found using the following equations:

ux = ucosθ = 10cos60 = 5 m/s
uy = usinθ = 10sin60 = 8.7 m/s

The horizontal motion of this projectile is uniform with a constant velocity of 5 m/s. The vertical motion of this projectile can be described by the same equation as before:

y = uyt - 1/2gt^2

where uy is the initial vertical velocity, t is the time, and g is the acceleration due to gravity. In this case, the initial vertical velocity is 8.7 m/s, so we have:

y = 8.7t - 1/2gt^2

After 1 second, the vertical displacement of this projectile is:

y = 8.7(1) - 1/2(9.8)(1)^2 = 3.6 m

Distance between the two projectiles:
The distance between the two projectiles can be found by calculating the horizontal displacement of the projectile launched at 60 degrees. The horizontal displacement can be found using the following equation:

x = uxt = 5(1) = 5 m

Therefore, after 1 second, the distance between the two projectiles is:

distance = x^2 + (y1 - y2)^2 = 5^2 + (5.1 - 3.6)^2 = 5.6 m

Therefore, the distance between the two projectiles after 1 second of their throw is 5.6 m.

1 wrote both velocities in vector form
2 find relative velocity between these two
3 take magnitude of this vector
4 use distance= speed*time

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