A body is thrown verticarry upwards with an initial velocity of 12m/s.(taking G to be 10m/s^2)
a)how far does it go during this first second?
b)find the maximum height it reaches.
c)how long does it remain in the air?
a) after 1 s the vertical velocity is 2 m/s
... so the average velocity is 7 m/s for the 1st sec
b) similar to a)
... the object reaches its peak after 1.2 s
... the average velocity is 6 m/s
c) time up equals time down
... 1.2 + 1.2
1.2+1.2=2.4
To solve this problem, we can use the kinematic equations of motion for motion in a vertical direction under constant acceleration.
Let's break down the problem and find the answers step by step:
a) How far does the body go during the first second?
We need to find the vertical displacement (distance) covered in the first second. To do that, we'll use the formula:
displacement (s) = initial velocity (u) × time (t) + (1/2) × acceleration (a) × time squared (t^2)
Given:
- Initial velocity (u) = 12 m/s (upwards)
- Time (t) = 1 second
- Acceleration (a) = -10 m/s^2 (since the acceleration due to gravity is downward)
Plugging in these values, we can calculate the distance covered during the first second:
s = (12 m/s) × (1 s) + (1/2) × (-10 m/s^2) × (1 s)^2
s = 12 m + (-5 m)
s = 7 m
Therefore, the body goes 7 meters during the first second.
b) Find the maximum height it reaches.
To find the maximum height, we need to determine the time it takes to reach that height. At the maximum height, the vertical velocity becomes zero (since the body momentarily stops before reversing its direction). We can use the formula:
final velocity (v) = initial velocity (u) + acceleration (a) × time (t)
Given:
- Initial velocity (u) = 12 m/s (upwards)
- Final velocity (v) = 0 m/s (at the highest point)
- Acceleration (a) = -10 m/s^2 (since the acceleration due to gravity is downward)
Plugging in these values and solving for time:
0 m/s = 12 m/s + (-10 m/s^2) × t
-12 m/s = -10 m/s^2 × t
t = -12 m/s / -10 m/s^2
t = 1.2 s
So, it takes 1.2 seconds to reach the maximum height.
To find the maximum height (h), we can use the formula:
h = initial velocity (u) × time (t) + (1/2) × acceleration (a) × time squared (t^2)
Plugging in the values:
h = (12 m/s) × (1.2 s) + (1/2) × (-10 m/s^2) × (1.2 s)^2
h = 14.4 m - 7.2 m
h = 7.2 m
Therefore, the maximum height reached by the body is 7.2 meters.
c) How long does it remain in the air?
Since the body is thrown upwards and eventually falls back to the ground, the total time in the air will be double the time it takes to reach the maximum height. Thus, the body remains in the air for:
Total time = 2 × time to reach the maximum height
Total time = 2 × 1.2 s
Total time = 2.4 s
Therefore, the body remains in the air for 2.4 seconds.