Range of a projectile

Resolving the velocity into components when given its magnitude and projection angleSuppose a ball is kicked upward with an initial velocity of 10 m/s at an angle of 30° from the horizontal. How far does it travel before it hits the ground? At what angle of elevation should you kick the ball for it to reach the maximum distance? Both questions are answered by applying the equations of motion in two dimensions. The equations of motion are the same as before (equations 6.6 and 6.7), except that we are now given the magnitude of the initial velocity v0 instead of the individual velocity components. Using the projection angle, we get the initial individual velocity components of vx0 = v0 cos θ and vy0 = v0 sin θ, leading to the modified equations given below. Read the text aloud
(6.8) x= v 0 t cosθ v x = v 0 cosθ
x  = position in x-direction (m)
v0  = initial speed (m/s)
θ  = projection angle (degrees)
t  = time (s)
vx  = velocity component in
x-direction (m/s)
Projectile motion
x-coordinate equations
(6.9) y= v 0 tsinθ 1 2 g t 2 v y = v 0 sinθgt
y  = position in y-direction (m)
v0  = initial speed (m/s)
θ  = projection angle (degrees)
t  = time (s)
vy  = velocity component in
y-direction (m/s)
Projectile motion
y-coordinate equations
The range of a projectile is the horizontal distance it covers before hitting the ground. That means x = v0t cos θ is the equation we want to solve. The initial speed v0 is known, and the angle θ is also known, while the time t is not known. So we use the y equations to solve for time t. The y-component equation for position is a quadratic, which means there are two solutions. Factoring the equation gives the following. Read the text aloud
Solving <i>y</i>-component of equation for time
The first solution is for t = 0; when the ball is first kicked at the start. The second solution is the one we want because it describes the second time the ball touches the ground. Solving the second solution for time yields t = 2v0 sin θ0 / g. Substituting this into x = v0t cos θ yields the range equation: Read the text aloud
(6.10) x= 2 v 0 2 sinθcosθ g
x  = range (m)
v0  = initial speed (m/s)
θ  = launch angle relative to horizontal
g  = acceleration of gravity (9.8 m/s2)
Range equation
What is the time for the ball to return to the ground when it is thrown directly upward at 18 m/s? Show

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