Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 40 degrees. It reaches the maximum height of 30 m. The acceleration due to gravity is 30 m/s2. Part 1. Find the time t the projectile spends in the air.

2.83 s

Explanation:

Let v be the initial launching velocity of the projectile. This can be split into 2 components:

- Horizontal component

- Vertical component

For the vertical motion, only affected by vertical velocity and gravitational acceleration g = 30 m/s2. As the object reaches maximum height, we can use the following equation of motion to find out the initial velocity of the projectile:

where v = 0 m/s is the final velocity of the object when it stops at the max height, is the initial vertical velocity of the object when it start, a = -30 m/s2 is the deceleration of the can, and is the maximum distance traveled.

Then we can also apply the following equation of motion to solve for the time it takes to reach maximum height of 30m

As we have a case of . The equation of t can be rewritten as

Since it would take another t seconds to fall down and hit the ground. The total time the projectile would spend on air is 2*1.415 = 2.83 s

car a and b are moving in opposite directions

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f=ma newton

f=100x2.5=250n