Question

A projectile is launched from ground level with an initial speed of 41.5 m/s at an angle of 32.3° above the horizontal. It strikes a target in the air 2.22 seconds later. What are the horizontal and vertical distances from where the projectile was launched to where it hits the target?

Answer 1

To answer this question it must be taken into account that according to the kinematics equations for the position of an object that performs this type of movement is:

$X (t) = V_{0} Cos (\alpha ) t$

$Y (t) = V_{0} Sen (\alpha ) t - 0.5gt ^ 2$

Where:

X (t) is the position of the object in the horizontal direction as a function of time

Y (t) is the position of the object in the vertical direction as a function of time

α is the angle of inclination in which the object is thrown

$V_{0}$ is the initial velocity of the object

g is the acceleration of gravity

t is the time

We want to know what distance the projectile traveled in 2.22 seconds after launch.

Then it is:

$X (t) = 41.5Cos (32.3) (2.22) = 77.87 m$

$Y (t) = 41.5 Sin (32.3) (2.22) - 0.5 (9.8) (2.22) ^ 2 = 25.08 m$

 After 2.22 seconds, the projectile strikes with the target at a horizontal distance of 77.87 m and a vertical distance of 25.08 m.