The ball would take sqrt(2H/g) = 7.00 s to hit the ground, but will reach the building in 720/120 = 6.00 s
So it will not hit the ground first, and will arrive in exactly 6.00 s.
A cannonball is fired horizontally from a 240 {m}-high cliff toward a vertical building 720 {m} away with an initial velocity of 120 {m/s}.
When the ball will hit the building?
in
a. 7.00 {s}
b. 6.00 {s}
c. less than 6.00 {s}
d. more than 7.00 {s}
2 answers
y=h=120m t=d/v y=1/2 g t^2 t=√(240/9.80s^2 ) t=440m/(110 m/s) t^2=2y/g
x=d=440m t=4s t=√(2y/g)
V_0=110 m/s 4s. to hit the building t=√((2(120m))/(9.80 m/s^2 ))
t=√(24.5s^2 )
t= ? To hit the ground in t=4.95s
x=d=440m t=4s t=√(2y/g)
V_0=110 m/s 4s. to hit the building t=√((2(120m))/(9.80 m/s^2 ))
t=√(24.5s^2 )
t= ? To hit the ground in t=4.95s
1 answer