Question

100 No ue busung une, who would win the race? (8 points? Com 2. A projectile is launched from ground level and at an angle of 53º. After 1.25 s, it just clears the comer of a building that is 30 m tall and 90 m wide. S tobom O b koostoot to o no ho samost 9 0 ML With what velocity was the projectile launched? (10 points) . How far from the launching point is the base of the building? (5 points) C. Will the projectile land on top of the building or on the ground on the other side of the building? Give the x-coordinate of the point where it lands. (10 points)

I know the answer to a) is 37.7 m/s and the answer to b) is 28.36 m

For c) here how I solve it.

Let’s find the range of the projectile:

that would be:

R = Vo ^2 /g * sin (2 theta)

= 37.7^2 / 9.8 * sin (2 x 53 degree)

= 139.41 m

Since 139.41m is longer than 28.36m + 90mm, it lands on the on the other side of the building.

Why is it not correct? please explain in details.

Answer 1

Usinssº 53 UCOS53" after 1025s displacement in y-direction ca is 3om. Y = uyt - 2 gt ² 30 U siassº (1.25) - 44.8) (1.25)? 3 30 = 0 (471-25) - ļ (2.8) (1-5626) 30 EU - 7.656 U - 30/7. 6S 6 ) U= 37 654 U=37.7 mis Y-937 - žom 2= Dict x = 010553') x1-25) u gom - 26 = 137.7)Cos53) (1-25) - 139.41m =R - 2=28.27 m, Yet's find out when horizontal displacement is Groom what is vertice displacement (gory)= Uxt & (90+28.27) = (37.7) Cas53) + o 4 = 5.23 seco Y = Uyt +4 Qyt > y = (37.7) sin53°(5023) -1 9.8) (5:25] Y = 157.736- 134.029 y = 23.706 m this shows that the expected - projectile path (trajactory is passing through through the the buildings building so it is not practically possible for it to cross the buildin