Question

How do I get the acceleration? also what is the formula i need to use to find the other velocity values?

Fill in the data table below for a projectile that is fired off of a building 57° BELOW the horizontal with a speed of 46 m/s. Take up to be positive. Ignore air drag. 0 1 2 (s) 25.0534 vх (m/s) 46 ax (m/s) -38.57885 Vy (m/s) ay (m/s2)

Answer 1

Initial velocity of projectile

1 = 46 m/s

As the projectile is fired below horizontal with angle $\theta=57\degree$

Initial horizontal velocity is

U u cos 0 = 46 cos(-57°) 46 cos(-57°) = 25.053 m

Initial vertical velocity is $u_y=u\sin\theta=46\sin(-57\degree)=-38.579\,m/s$

As the projectile moves in the air, the horizontal component of velocity does not change with time
Vr = U 25.053 m/s
as there is no acceleration is the horizontal direction $a_x=0\,m/s^2$ ( at all times) and vertical component of velocity changes with time $v_y=u_y-gt$ as the acceleration in the vertical direction is $a_y=-g=-9.81\,m/s^2$ (at all times)

At time $t=0\,s$

$v_y=-38.579-9.81*0=-38.579\,m/s$ and $a_y=-9.81\,m/s^2$

At time $t=1\,s$

$v_y=-38.579-9.81*1=-48.389\,m/s$ and $a_y=-9.81\,m/s^2$

At time $t=2\,s$

$v_y=-38.579-9.81*2=-58.199\,m/s$  and $a_y=-9.81\,m/s^2$ ​​​​​​​

At time $t=3\,s$

$v_y=-38.579-9.81*3=-68.009\,m/s$  and $a_y=-9.81\,m/s^2$ ​​​​​​​

At time $t=4\,s$

$v_y=-38.579-9.81*4=-77.819\,m/s$ and $a_y=-9.81\,m/s^2$ ​​​​​​​

t(s)	0	1	2	3	4
vx (m/s)	25.053	25.053	25.053	25.053	25.053
ax(m/s2 )	0	0	0	0	0
vy(m/s)	-38.579	-48.389	-58.199	-68.009	-77.819
ay(m/s2)	-9.81	-9.81	-9.81	-9.81	-9.81