A projectile is launched with an initial speed of 59.0 m/s at an angle of 32.0° above the horizontal. The projectile lands on a hillside 3.80 s later. Neglect air friction. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)

Question

(a) What is the projectile's velocity at the highest point of its trajectory?
magnitude ? m/s

direction ? ° counterclockwise from the +x-axis

(b) What is the straight-line distance from where the projectile was launched to where it hits its target? in m

Answer 1

in the vertical direction
... h = 1/2 g t^2 + 59 sin(32º) t

in the horizontal direction
... d = 59 cos(32º) t

use the axis of symmetry of the parabola to find the time for max height

plug the time into the vertical equation to find the max height

the initial energy minus the gravitational potential at max height gives the energy (velocity) at max height

the projectile has no vertical velocity at max height, so the direction is 0º

plug the flight time into the vertical equation to find the impact height

plug the flight time into the horizontal equation to find the horizontal distance

use Pythagoras to find the straight line distance