A projectile is shot upward from the earth with an initial velocity of 320 ft/sec.
(a) What is the velocity after 5 secs.? ft/sec.
(b) What is the acceleration after 3 secs.? ft/sec^2.
a. V = Vo + g*t = 320 - 32*5 = 160 Ft/s.
To solve this problem, we need to use the equations of motion for projectiles. These equations describe the motion of an object in freefall under the influence of gravity.
The equations of motion for the vertical direction are:
1. Velocity equation: v = u + at
2. Displacement equation: s = ut + (1/2)at^2
3. Acceleration equation: a = -g
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = displacement
g = acceleration due to gravity (approximately 32 ft/sec^2)
Let's solve the problem step by step.
(a) What is the velocity after 5 secs.?
Using the velocity equation:
v = u + at
Given:
u = 320 ft/sec
t = 5 secs
a = -g = -32 ft/sec^2
Substituting the values into the equation:
v = 320 + (-32)(5)
v = 320 - 160
v = 160 ft/sec
Therefore, the velocity after 5 seconds is 160 ft/sec.
(b) What is the acceleration after 3 secs.?
Using the acceleration equation:
a = -g
Given:
g = 32 ft/sec^2
Substituting the value into the equation:
a = -32 ft/sec^2
Therefore, the acceleration after 3 seconds is -32 ft/sec^2.
To calculate the velocity and acceleration of a projectile after a certain time, we need to use the equations of motion.
(a) To find the velocity after 5 seconds, we can use the equation of motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 320 ft/sec (initial velocity)
t = 5 sec (time)
We need to find the final velocity (v):
Using the equation:
v = u + at
v = 320 + (acceleration * time)
We don't have the acceleration value given in the question. The acceleration acting on a projectile when shot upward near the Earth's surface is due to gravity and is approximately -32 ft/sec^2 (negative sign indicates that it acts in the opposite direction to the initial velocity).
So, substituting the values:
v = 320 + (-32 * 5)
v = 320 - 160
v = 160 ft/sec
Therefore, the velocity of the projectile after 5 seconds is 160 ft/sec.
(b) To find the acceleration after 3 seconds, we know that the acceleration due to gravity near the Earth's surface is approximately -32 ft/sec^2. This value remains constant throughout the motion.
So, the acceleration after 3 seconds is also -32 ft/sec^2.
Therefore, the acceleration after 3 seconds is -32 ft/sec^2.