Agolfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is the ball’s maximum height? (hint: at the top of its flight, the ball’s vertical velocity component will be zero.

the horizontal displacement can be expressed as:

Δx=vxt

Height is expressed as Δy=vyt−12gt2

The time it would hit the maximum height is expressed as:

0=vy−gt→t=vyg

The time that the projectile reaches the ground is twice the time it takes the projectile to reach its maximum height. This is expressed as

t=2vyg

to find the time of the projectile to reach the maximum height, we have the formula

r=vx2vyg

From trigonometry,

vx=cos(θ)vy=sin(θ)

r=2cos(θ)sin(θ)g

We know from our trigonometric identities that

sin(2θ)=2sin(θ)cos(θ) , so it would become

r=v2gsin(2θ)

ymax=v2yg−12g(vyg)2=v2yg−12v2yg=v2sin2(θ)2g

velocity is R=301.5

θ

=25 g=10

m/s2

V(0)=? Y(max)=?

R=R=v2∗sin(2θ)/g

Substitute the answers  in the equation v ll,  that would be  V^2=3935m/s we also know that at the highest point V(y)=0 so we v can write down:

V(y)2−V(0)(y)2=−2gΔy

we also know that

V(y)=V(0)∗sinθ

V(y)2−[V(0)sinθ]2=−2gΔy

y :35m

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step-by-step explanation:

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