A projectile fired from a gun has initial horizontal andvertical components of velocity equal to 30m/s and 40m/srespectively.
Determine the initial speed of the projectile.
Approximately how long does it take the projectile to reachthe highest point in its trajectory and the max heightH of the projectile?
What is the speed of the projectile when it is at the highestpoint in its trajectory and what is the acceleration of theprojectile when it reaches its maximum height?
the answers for the acceleration of the projectile when itreaches it maximum height.
9.8 m/s2 downward,
4.9 m/s2 downward,
less than 9.8 m/s2 and non-zero,
its magnitude is 9.8 m/s2; and its direction ischanging
What is the Range of the projectile.
I tried these equations to figure out this problem.
1) the square root of x2 + y2 =the square root of 302 + 402 = the squareroot of 2500 = 50m/s
2) 50m/s/9.8m/s2 = 5.1 s for thecomplete time. Then divide by 1/2 to get the time to theheighest point. 5.1s/2 2.55 s = 2.55s
3) (50m/s)2 = ( square root of 2 time9.8/h)2 = 2500/(2 times 9.8 ) = h 127.55m =h
4) r=VxT 30m/s times 5.1s = 153m
5) 2.5s = 2Vy/9.8m/s 9.8m/s2 times 2.5s =2Vy 24.99m/s = 2Vy 24.99/2=Vy 12.99 = Vy
6)I chose 9.8 m/s2 downward. But I think it iszero m/s2