a body is projected horizontally from height of 78.4 m with a velocity 10 m/s . its velocity after 3 sec
To find the velocity of the body after 3 seconds, we can use the equations of motion.
Since the body is projected horizontally, its initial vertical velocity is zero, and the only force acting on it is gravity. Thus, the motion in the vertical direction is solely due to the acceleration due to gravity.
First, let's find the time it takes for the body to hit the ground using the equation:
\[h = \frac{1}{2} g t^2\]
where h is the initial height, g is the acceleration due to gravity, and t is the time taken.
Plugging in the given values:
\[78.4 = \frac{1}{2} \cdot 9.8 \cdot t^2\]
Simplifying the equation:
\[156.8 = 4.9t^2\]
Dividing both sides of the equation by 4.9:
\[t^2 = \frac{156.8}{4.9}\]
\[t^2 \approx 32\]
Taking the square root of both sides:
\[t \approx \sqrt{32} \approx 5.66\]
So, it takes approximately 5.66 seconds for the body to hit the ground.
Now, let's find the horizontal distance traveled by the body in 3 seconds using the equation:
\[d = v \cdot t\]
where d is the distance, v is the initial velocity, and t is the time taken.
Plugging in the given values:
\[d = 10 \cdot 3\]
\[d = 30\]
Therefore, the horizontal distance traveled by the body in 3 seconds is 30 meters.
Since there are no horizontal forces acting on the body, the horizontal velocity remains constant throughout the motion. Thus, the velocity after 3 seconds is the same as the initial horizontal velocity, which is \(10 \, \text{m/s}\).
To find the velocity of the body after 3 seconds, we can use the concept of projectile motion.
When a body is projected horizontally, the vertical motion of the body is influenced by the force of gravity, while the horizontal motion is not affected by gravity.
Given:
Height (h) = 78.4 m
Velocity (u) = 10 m/s
Time (t) = 3 sec
First, we need to calculate the initial vertical velocity component (v) of the body:
v = u + gt
v = 10 m/s + (9.8 m/s²) * 3 sec
v = 10 m/s + 29.4 m/s
v = 39.4 m/s
Since the vertical motion is only influenced by gravity, the vertical velocity (v) remains constant throughout the motion. Therefore, the velocity of the body after 3 seconds is 39.4 m/s horizontally.
First, check that it hasn't yet hit the ground.
Finding it has not, then get the resultant of the horizontal and vertical velocities
√(10^2 + (9.8*3)^2)
at an angle of 71° below the horizontal
u = 10 m/s forever
v = -gt = -9.81 (3) = - 29.4 m/s
check if it hit the ground
d = (1/2) g t^2 = 4.9 * 9 = 44.1 so still in the air
speed = sqrt (10^2 + 29.4^2)