The rock was thrown horizontally, so u, the initial vertical velocity = 0
Use equations of motion:
u = initial vertical velocity = 0
S = vertical distance travelled, m
t = time, s
g = acceleration due to gravity, 9.81 m/s²
S = ut + (1/2)at²
Solve for S.
A boy standing on top of a hill throws a stone horizontally. The stone hits the ground at the foot of the hill 2.5 s later. How high is the hill ?
3 answers
v1y = 0
a = -9.8 m/s
t = 2.5s
d = ?
d = v1t + 1/2(a)(t)^2
d = 0 + 1/2 (-9.8)(2.5)^2
d = -30.625
a = -9.8 m/s
t = 2.5s
d = ?
d = v1t + 1/2(a)(t)^2
d = 0 + 1/2 (-9.8)(2.5)^2
d = -30.625
The boy is standing on top of a hill throws a stone horizontally. The stone hits the ground at the foot of the hill 2.5 seconds later. How high the hill?