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 ?
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
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.
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?
To find the height of the hill, we can use the equation of motion for a projectile that moves horizontally and vertically simultaneously. The equation is:
h = (1/2) * g * t²
Where:
h is the height of the hill,
g is the acceleration due to gravity (approximately 9.8 m/s²),
t is the time it takes for the stone to hit the ground (2.5 s in this case).
Now, let's substitute the values into the equation and solve for h:
h = (1/2) * 9.8 m/s² * (2.5 s)²
h = (1/2) * 9.8 m/s² * 6.25 s²
h = 4.9 m/s² * 6.25 s²
h = 30.625 m
Therefore, the hill is approximately 30.625 meters high.