a ball leaves the club of a golfer at a speed of 33m/s at an angle of 37 with respect to the horizontal. what is its speed when it reach its maximum height?
Since the ball stops at its' maximum
height, the velocity is zero.
To find the speed of the ball when it reaches its maximum height, we need to analyze the motion of the ball.
Step 1: Split the initial velocity into horizontal and vertical components.
The horizontal component remains constant throughout the motion, while the vertical component changes due to the effect of gravity.
Given:
Initial speed, v = 33 m/s
Launch angle, θ = 37 degrees
The horizontal component (v_x) is given by:
v_x = v * cos(θ)
The vertical component (v_y) is given by:
v_y = v * sin(θ)
Step 2: Calculate the time taken to reach the maximum height.
At the maximum height, the vertical component of velocity becomes zero. In projectile motion, the time taken to reach the maximum height is half of the total time of flight.
Using the equation:
v_y = u_y + a * t
where,
v_y = vertical component of velocity (0 m/s at maximum height)
u_y = initial vertical component of velocity (v * sin(θ))
a = acceleration due to gravity (-9.8 m/s²)
t = time taken to reach the maximum height
0 = v * sin(θ) - 9.8 * t
Solving for t:
t = v * sin(θ) / 9.8
Step 3: Find the speed at maximum height.
To find the speed at maximum height, we need to evaluate the vertical component of velocity (v_y) using the time calculated in Step 2.
v_y = u_y + a * t
v_y = v * sin(θ) - 9.8 * t
Substituting the value of t from the previous step:
v_y = v * sin(θ) - 9.8 * (v * sin(θ) / 9.8)
v_y = v * sin(θ) - v * sin(θ)
v_y = 0 m/s
Therefore, the speed of the ball when it reaches its maximum height is 0 m/s.