Question

Two projectiles are launched into the air. The first projectile has a y-component of its initial velocity of 10 m/s and an x-component of its initial velocity of 10 m/s. (This is equivalent to an initial speed of 14.2 m/s at a 45 degree angle.) The second projectile also has a y-component of its initial velocity of 10 m/s but an x-component of 5 m/s for its initial velocity. (This is equivalent to an initial speed of 11.2 m/s at a 63 degree angle.)

First, if both projectiles were launched at the same time on a horizontal surface, which one would land first, or would they land at the same time?

Second, which projectile would travel farther before landing, or would they travel the same distance?

Third, which projectile would reach the greatest maximum height off the ground, or would the maximum height be the same?

Answer 1

a)

from kinematic equation

Y=Y_o+V_oyt-(1/2)gt²

When projectile lands on ground Y=0

0=0+V_oyt-(1/2)gt²

t=sqrt[2*V_oy/g]

for first projectile

t₁=sqrt(2*10/9.8) =1.43 s

for second projectile

t₂=sqrt(2*10/9.8)=1.43 s

Therefore both projectile would land at the same time.

b)

Horizontal distance traveled by each projectile

D=V_oxt

For each projectile

D₁=10*1.43=14.3 m

D₂=5*1.43=7.15 m

Therefore Projectile 1 would travel farther before landing.

C)

From Kinematic equation

V_fy²=V_oy²-2gh

At maximum height h=H_max ,V_fy=0

=>0=V_oy²-2gH_max

H_max =V_oy²/2g

For each projectile

H_Max,1=10²/2*9.8 =5.1 m

H_Max,2=10²/2*9.8 =5.1 m

Therefore Both projectile would reach the same maximum height.