A ball projected with an initial velocity u at an angle tita to the horizontal cover a horizontal range of 26m after 5sec. Cal. The value of u and tita? g=10ms-1
well I call initial velocity horizontal component u and vertical component Vi
but your way
horizontal velocity = z = u cos theta
initial vertical velocity = Vi = u sin theta
z = 26/5 = u cos theta = 5.2
now it tskes 5 seconds to go up and down
so it takes 2.5 seconds to go up
v = Vi - 10 t
0 = Vi - 10 t
so
t = u sin theta/10 = 2.5
so
u sin theta = 25
therefore
u sin theta/u cos thets = 25/5.2
or
tan theta = 4.81
theta = 78.3 degrees
and u = 25/sin 78.3
To find the value of the initial velocity (u) and the angle (theta), we can use the equations of motion for projectile motion. Let's break down the problem step by step.
Step 1: Determine the horizontal component of velocity (ux) and the vertical component of velocity (uy).
Given:
Horizontal range (R) = 26m
Time of flight (t) = 5s
Acceleration due to gravity (g) = 10 m/s^2
The horizontal component of velocity (ux) remains constant throughout the motion, while the vertical component of velocity (uy) changes due to the effect of gravity.
We know that the horizontal range (R) is given by:
R = ux * t
26m = ux * 5s
ux = 26m / 5s
ux = 5.2 m/s
Step 2: Use the equation to calculate the vertical component of velocity at any time (uy).
The equation for the vertical component of velocity is:
uy = u * sin(theta) - g * t
At the highest point of the projectile's trajectory, the vertical velocity component becomes zero. It occurs when the projectile reaches the highest point of the trajectory halfway through the time of flight.
Using t = 2.5s (half of the total time of flight), and uy = 0, we can calculate u * sin(theta).
0 = u * sin(theta) - g * 2.5s
Step 3: Calculate the initial velocity (u) and the angle (theta).
Given that g = 10m/s^2, we can substitute the values into the equation:
0 = u * sin(theta) - 10m/s^2 * 2.5s
u * sin(theta) = 25m/s
Divide the equation by sin(theta):
u = 25m/s / sin(theta)
Now, we can use the range equation to substitute in the value of u:
26m = (25m/s / sin(theta)) * 5s
Rearrange the equation to solve for sin(theta):
sin(theta) = (25m/s * 5s) / 26m
sin(theta) = 4.8077
To get the value of theta, take the inverse sine (or arcsine) of sin(theta):
theta = arcsin(4.8077)
However, there is no real value for arcsin(4.8077). This means that the given values do not result in a valid trajectory. Please recheck the values or the question details.
Note: If there was a valid solution, you could calculate the angle (theta) from the inverse sine function and the initial velocity (u) using u = 25m/s / sin(theta).