A projectile is launched straight upwards at 45 m/s. Three seconds later, its velocity is
A. 75 m/s
B. 30 m/s
C. 15 m/s
D. zero
15
v=45-g*time
To find the velocity of the projectile 3 seconds after it was launched, we can use the formula for projectile motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity is 45 m/s, the acceleration is due to gravity (-9.8 m/s^2), and the time is 3 seconds. Therefore, we can substitute these values into the formula:
v = 45 + (-9.8)(3)
v = 45 - 29.4
v = 15.6 m/s
Therefore, the velocity of the projectile 3 seconds after it was launched is 15.6 m/s.
The correct answer is C. 15 m/s.
To find the velocity of the projectile three seconds after it is launched, we need to consider the effect of gravity. The acceleration due to gravity is approximately 9.8 m/s², acting downward.
Since the projectile is launched straight upwards, its initial velocity is positive (+45 m/s). After three seconds, the only force acting on the projectile is gravity, which causes it to decelerate. We can use the equation:
final velocity = initial velocity + (acceleration due to gravity * time)
Plugging in the values:
final velocity = 45 m/s + (-9.8 m/s² * 3 s)
final velocity = 45 m/s - 29.4 m/s
final velocity = 15.6 m/s
Therefore, three seconds later, the velocity of the projectile is approximately 15.6 m/s. The correct answer is C. 15 m/s.