Question

1. four projectiles, a, b, c, and d, were launched from, and returned to, level ground. the data table below shows the initial horizontal speed, initial vertical speed, and time of flight for each projectile. which projectile traveled the greatest horizontal distance? neglect friction.

table 1: data table

projectile	initial horizontal speed (m/s)	initial vertical speed (m/s)	time of flight (s)
b	60.0	19.6	4.00
c	50.0	24.5	5.00
d	80.0	19.6	4.00

Explanation:

Step1: Recall horizontal - distance formula

The horizontal distance $d_x$ of a projectile is given by $d_x = v_{0x}t$, where $v_{0x}$ is the initial horizontal speed and $t$ is the time of flight.

Step2: Calculate horizontal distance for projectile A

For projectile A, $v_{0x}=40.0\ m/s$ and $t = 6.00\ s$. So $d_{xA}=v_{0x}t=40.0\times6.00 = 240\ m$.

Step3: Calculate horizontal distance for projectile B

For projectile B, $v_{0x}=60.0\ m/s$ and $t = 4.00\ s$. So $d_{xB}=v_{0x}t=60.0\times4.00=240\ m$.

Step4: Calculate horizontal distance for projectile C

For projectile C, $v_{0x}=50.0\ m/s$ and $t = 5.00\ s$. So $d_{xC}=v_{0x}t=50.0\times5.00 = 250\ m$.

Step5: Calculate horizontal distance for projectile D

For projectile D, $v_{0x}=80.0\ m/s$ and $t = 4.00\ s$. So $d_{xD}=v_{0x}t=80.0\times4.00=320\ m$.

Answer:

Projectile D