Question

A projectile is launched with an initial speed of 45.0 m/s at an angle of 40.0° above the horizontal. The projectile lands on a hillside 3.25 s later. Neglect air friction. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)

(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?
?  m

Answer 1

V_o = initial velocity = 45 m/s

= angle = 40

initial Velocity along x-direction = V_ox = V_o Cos = 45 Cos40 = 34.5 m/s

initial Velocity along y-direction = V_oy = V_o Sin = 45 Sin40 = 28.9 m/s

t = time taken = 3.25 sec

a_y = acceleration in vertical direction = -9.8 m/s²

a_x = acceleration in horizontal direction = 0

a)

velocity after time 't' along x-direction is given as

V_fx = V_ox + a_x t
V_fx = 34.5 + 0 x 3.25 = 34.5 m/s

velocity after time 't' along y-direction is given as

V_fy = V_oy + a_y t
V_fy = 28.9 + (-9.8) x 3.25 = - 2.95 m/s

net velocity is given as

V_f = sqrt ((V_fx )² + (Vfy)² )

V_f = sqrt ((34.5)² + (-2.95)² )

V_f = 34.63 m/s

= tan^-1 (V_fy /V_fx) = tan^-1 (2.95 / 34.5) = 4.89 below X-axis

counterclockwise , angle = 360 - 4.89 = 355.11

b)

displacement in Y-direction is given as

Y = V_oy t + (0.5) a_y t²

Y = 28.9 x 3.25 + (0.5) (-9.8) (3.25)²

Y = 42.2 m

displacement in X-direction is given as

X = V_ox t = 34.5 x 3.25 = 112.13 m

Net distance = sqrt (X² + Y²) = sqrt (42.2² + 112.13²) = 119.8 m