If a projectile is fired straight up at a speed of 13m/s. How long does it take to reach the top of its motion? If the acceleration due to gravity is 9.8m/ s motion (squared) answer in units of s
V = Vo + gt.
t = (V-Vo)/g = (0-13)/-9.8 = 1.33 s.
To find the time it takes for the projectile to reach the top of its motion, we can use the following formula:
time = (final velocity - initial velocity) / acceleration
Given:
Initial velocity (u) = 13 m/s (positive, as it is fired straight up)
Acceleration due to gravity (g) = -9.8 m/s^2 (negative, as gravity pulls the projectile downward)
Plug these values into the formula:
time = (0 - 13) / (-9.8)
= -13 / (-9.8)
≈ 1.32 seconds.
Therefore, it takes approximately 1.32 seconds to reach the top of its motion.
To find the time it takes for the projectile to reach the top of its motion, we can use the kinematic equation:
Vf = Vi + at
where:
- Vf is the final velocity,
- Vi is the initial velocity,
- a is the acceleration, and
- t is the time.
In this case, the projectile's initial velocity (Vi) is 13 m/s and it is fired straight up, so we can assume the acceleration due to gravity (a) is -9.8 m/s^2 (negative sign indicates that the acceleration is pointing downward).
At the top of the projectile's motion, the final velocity (Vf) becomes 0 m/s since it momentarily comes to a stop before falling back down. Therefore, our equation becomes:
0 = 13 m/s + (-9.8 m/s^2)t
Now, we can solve for t:
-9.8t = -13
Dividing both sides of the equation by -9.8, we get:
t = -13 / -9.8
t ≈ 1.33 seconds
Thus, it takes approximately 1.33 seconds for the projectile to reach the top of its motion.