What is the rocket’s maximum range?

Question

A soccer ball is kicked with an initial velocity of 45 km/h at an angle of 55° above the horizontal. What is the rocket’s maximum range?

Answer 1

Hi, it's pretty good to ask this question. The only thing you have to do is just to plug these given numbers in our equation that we then use to find the range directly, which is:

R = vi² sin(2θi) / g

R is the range.

v_i is the initial velocity.

θ_i is the initial angle that you started your shooting with.

g is the constant of gravity.

All you want here is just to substitute by their values in our equation.

The final result is 14.98 meter

Be careful: I changed the unit from km/h to m/s.

Answer 2

In physics, for such a problem, we should realize first and foremost that the SI units should be in meters per second. To do this conversion, we would have to set up 45 km/hr * 1000 m/1km * 1 hr/3600 seconds.

We now should have the initial velocity to be 45 meters/second at an angle of 55 degrees

The key to solving this problem would be noting that when this soccer ball is kicked, its velocity would be 12.5 meters per second OVERALL - we would need to break the balls velocity into the x component and y component.

The x component in projectile motion is always the VelocityInitial * Cos(Theta) --> 12.5 m/s Cos (55) = 7.17 m/s is the speed in the x direction. In the x direction, there is no NET FORCE acting, it is only in the y direction that gravity acts. Thus, the range (distance traveled in the x-direction) would be Distance = Vx * t of travel.

However, the time that the ball travels depends on the y component of the balls velocity which would be Velocity initial * Sin(theta) = 10.24 m/s in the y direction. In the y direction, when the ball reaches its max height, it would have a velocity of 0 m/s in the y direction momentarily (peak height). We can take advantage of this and say that the Vf = Vo + at ---> (a = acceleration due to gravity in y direction)

0 = 10.24 + (-9.8)*t -- solving for the time it takes to reach the peak height, we get 1.04 seconds to reach the max height. In projectile motion of such cases, this is also half of the time for the full trip - it takes 1.04 seconds to go from the ground to the max height, and another 1.04 seconds to go from the max height back to the ground. Thus, the total travel time for the soccer ball is roughly 2.08 seconds. Going back to our x direction where there is NO acceleration due to gravity, we can use Distance = Vx * t -->

Distance = 7.17 meters/second * 2.08 seconds => which gives us our Range of ~ 14.99 Meters

(if they wanted the answer in Km -- we would have 14.99 meters * 10^-3 km/m --> 0.01408 km)