A golf ball is hit a horizontal distance of exactly 300 m. What is the maximum height the golf ball reaches in the air if it is launched at an angle of 25 degrees to the ground?

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karton · Accepted Answer

To find the maximum height reached by the golf ball, we can use the equations of projectile motion. The horizontal and vertical motion of the golf ball can be treated independently.Given:- Horizontal distance traveled (range) = 300 m- Launch angle = 25 degreesWe can use the following equations:1. The horizontal component of the ball&#039;s velocity remains constant throughout its flight:vhorizontal=v·cos(θ)2. The vertical component of the ball&#039;s velocity changes due to the effect of gravity:vvertical=v·sin(θ)−g·t3. The time of flight, which is the total time the ball is in the air, can be calculated using:t=2vverticalg4. The maximum height reached by the ball can be found using:hmax=vvertical·t−12g·t2Given that we know the horizontal distance traveled is 300 m, we need to find the initial velocity v to solve for the maximum height.Using the range equation:300=vhorizontal·tSince the horizontal velocity vhorizontal=v·cos(θ), we can rearrange the equation to solve for v:v=300t·cos(θ)Substituting this value of v into the equation for the time of flight, we have:t=2vverticalg=2·(v·sin(θ))gNow we can substitute the expressions for v and t into the equation for the maximum height hmax:hmax=vvertical·t−12g·t2=(v·sin(θ)−g·t)·t−12g·t2Plugging in the values we know, we have:hmax=(300t·cos(θ)·sin(θ)−g·t)·t−12g·t2Now we can calculate the value of hmax by substituting the values of t, g, and θ:hmax=(3002·(v·sin(θ))g·cos(θ)·sin(θ)−g·2·(v·sin(θ))g)·2·(v·sin(θ))g−12g·(2·(v·sin(θ))g)2Simplifying this expression will give us the maximum height reached by the golf ball.

A golf ball is hit a horizontal distance

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