A projectile is shot at 42o with respect to the horizon in the direction of a th
ID: 2000849 • Letter: A
Question
A projectile is shot at 42o with respect to the horizon in the direction of a thin wall 60.0 m away. If the wall is 15.0 m high, then
a) What should be the initial velocity of the projectile to just clear the top of the wall?
b) If the thin wall is just next to a ditch 50.0 m deep as shown, how long will it take for the projectile to
hit the bottom and what will be the magnitude of its velocity at that point?
Adobe Reader File Edit View Window Help 01-Midterm 1 review.pdf 1AA Tools Sign Comment A projectile is shot at 42° with respect to the horizon in the direction of a thin wall 60.0 m away. If the wall is 15.0 m high, then 5. a) What should be the initial velocity of the projectile to just clear the top of the wall? 42°Explanation / Answer
a) suppose projectile reaches above the wall in time t.
so in time t it have to travel 60m in horizontal and more than 15 m in vertical
in this time.
in horizontal,
(vcos42)t = 60
vt = 60/cos42 = 80.74 ........(i)
in vertical,
h = uy*t - gt^2 /2
15 = (vsin42)t - (9.81t^2 /2 )
15 = 0.669vt - 4.905t^2
15 = 0.669(80.74) - 4.905t^2
t = 2.82 sec
v = 80.74/2.82 = 28.62 m/s ......Ans
b) ball have to 50m vertically downn from its initial position.
vertical displacement y = -50m
y = uy*t - gt^2 /2
-50 = (28.62 sin42 t ) - (4.905t^2)
4.905t^2 - 19.15t - 50 = 0
t = 5.69 sec
vx = 28.62cos42 = 21.27 m/s
vy = 19.15 - (9.81*5.69) = - 36.67 m/s
speed = sqrt(vx^2 + vy^2) = 42.39 m/s