Stuck on a Kinematics Problem?

The horizontal, 8.3 m/s, velocity is constant for the entire trip.

As the projectile rises to its maximum height, the vertical velocity decreases from the initial vertical velocity to 0 m/s at the rate of 9.8 m/s each second. As the projectile falls to the ground, the vertical velocity increases from 0 m/s to the final vertical velocity at the rate of 9.8 m/s each second. The distance the projectile rises is equal to the distance the projectile falls. The time for the upward trip is equal to the time for the downward trip. So, the initial vertical velocity is equal to the final vertical velocity.

Let’s determine the initial vertical velocity.

vf = vi – 9.8 * t

The vertical velocity is 4.6 m/s after 3.3 seconds

4.6 = vi – 9.8 * 3.3

vi = 4.6 + 9.8 * 3.3 = 36.94 m/s

The initial vertical velocity = 36.94 m/s. So, the final vertical velocity = 36.94 m/s

Since we know the horizontal velocity, let’s determine the magnitude and angle of the initial and final velocity.

Magnitude = (8.3^2 + 36.94^2)^0.5

The magnitude of the initial and final velocity is approximately 37.86 m/s

Tangent of angle above horizontal = 36.94 ÷ 8.3

The angle above horizontal is approximately 77.34˚

When the projectile reaches the maximum height, the vertical velocity is 0 m/s

vf = vi – 9.8 * t

0 = 36.94 – 9.8 * t

Time = 36.94 ÷ 9.8

This is the time for the upward trip. So this is the time for the downward trip.

For the downward trip, the vertical velocity increases from 0 m/s to the final vertical velocity at the rate of 9.8 m/s each second.

d = vi * t + ½ * 9.8 * t^2

d = ½ * 9.8 * (36.94 ÷ 9.8)^2

The distance the projectile falls is approximately 69.62 meter. So, the distance the projectile rises is approximately 69.62 meters. This is the maximum height.

The horizontal velocity is constant for the entire time. Horizontal distance = 8.3 * total time

Total time = Time up + Time down = 2 * 36.94 ÷ 9.8

Horizontal distance = 8.3 * 2 * 36.94 ÷ 9.8

The range is approximately 62.57 meters.

a) Determine the horizontal range of the projectile = 62.57 meters

b)Determine its maximum height above the ground = 69.62 meters

c)Determine the speed of motion just before the projectile strikes the ground = 37.86 m/s

d)Determine the angle of motion just before the projectile strikes the ground = 77.34˚ above horizontal

I hope this helps you understand how to solve this type of problem.

while uncertain, start up with the main undemanding of each and every of the kinematic relationships: distance = ordinary velocity X time traveled. making use of SUVAT notation, in math talk, it is S = Vavg T = 1300 meters the place the gap is in height. while A = consistent, then Vavg = (U + V)/2 the place V = U + AT is the cost after accelerating at A for T gadgets of time. So assuming the beginning velocity for the rocket is U = 0, then Vavg = V/2 = AT/2 the place V = 230 mps. Then T = S/Vavg = 1300 m/(230 mps/2) = 2600/230 = 11.3 sec. So from V = AT we've 230/11.3 = A = 20.35 m/s^2 ANS.