A particle is projected with a speed of 25 m/s at an angle of 40 degrees above the horizontal. "What is the time taken to reach the highest point of trajectory?"
Work with the vertical component of v0sin(θ) to get the required time, using the equation:
(-g)t=(v-v0)sin(θ)
To find the time taken to reach the highest point of the trajectory, you can use the kinematic equations of motion.
Step 1: Resolve the initial velocity into its horizontal and vertical components.
Given:
Initial speed (u) = 25 m/s
Angle (θ) = 40 degrees
The horizontal component (u_x) can be calculated using:
u_x = u * cos(θ)
The vertical component (u_y) can be calculated using:
u_y = u * sin(θ)
Step 2: Determine the time taken to reach the highest point.
When the particle reaches the highest point of its trajectory, its vertical velocity (v_y) becomes zero. We can use the kinematic equation for vertical displacement to find the time (t) taken to reach this point.
v_y = u_y + g * t (where g is the acceleration due to gravity, approximately 9.8 m/s^2)
0 = u_y + g * t
Solving for t:
t = -u_y / g
Substituting the values:
t = - (u * sin(θ)) / g
Step 3: Calculate the time.
Substitute the given values:
u = 25 m/s
θ = 40 degrees
g = 9.8 m/s^2
t = - (25 * sin(40)) / 9.8
Using a calculator, evaluate this expression to find the time taken to reach the highest point of the trajectory.