a body is thrown vertically upward return to the earth in 3sec.
(a)what was the initial velocity of the body
(b)what height did the body reach?
neglect the resistance of air.
a. hf=ho+vi*t-4.8t^2
hf,ho are zero, solve for vi
b. same equation,t=1.5
To solve this problem, we can use the equations of motion under constant acceleration. In this case, the acceleration is due to gravity and acts in the opposite direction to the motion of the body when it is thrown upwards.
(a) To find the initial velocity of the body, we can use the formula:
v = u + at
Where:
v = final velocity (which is 0 when the body reaches its highest point)
u = initial velocity
a = acceleration due to gravity (approximated as 9.8 m/s^2)
t = time taken (which is 3 seconds in this case)
Rearranging the formula, we get:
u = v - at
Since we know the final velocity is 0, we can substitute its value into the formula:
u = 0 - (9.8 m/s^2)(3 s)
u = -29.4 m/s
Therefore, the initial velocity of the body is -29.4 m/s (negative sign indicates upward motion).
(b) To find the maximum height the body reaches, we can use the kinematic equation:
s = ut + (1/2)at^2
Where:
s = distance traveled (height in this case)
u = initial velocity (-29.4 m/s)
a = acceleration due to gravity (-9.8 m/s^2, negative as it opposes the motion)
t = time taken (3 seconds)
Plugging in the values, we have:
s = (-29.4 m/s)(3 s) + (1/2)(-9.8 m/s^2)(3 s)^2
s = -88.2 m + (1/2)(-9.8 m/s^2)(9 s^2)
s = -88.2 m - 44.1 m
s = -132.3 m
The negative sign indicates that the body reached a height of 132.3 meters below its initial position.