A ball is thrown horizontally at 15 m/s from a clifftop 70 m above the sea level. Find:
a) the time to land;
s
b) the angle at which it hits the water;
° below the horizontal
c) the speed at which it hits the water.
m/s
To solve this problem, we can use the equations of motion.
a) The time to land:
We can use the vertical motion equation: h = ut + (1/2)gt^2, where h is the vertical displacement, u is the initial vertical velocity, g is the acceleration due to gravity (approximated as -9.8 m/s^2), and t is the time.
Given that the ball is thrown horizontally, the initial vertical velocity is zero, and the height of the cliff is 70 m. Plugging in these values into the equation, we get:
70 = 0 * t + (1/2)(-9.8)(t^2)
70 = -4.9t^2
Solving for t, we get:
t^2 = 70 / 4.9
t^2 = 14.29
t = √14.29
t ≈ 3.78 s
Therefore, the time it takes for the ball to land is approximately 3.78 seconds.
b) The angle at which it hits the water:
The angle can be found using the horizontal motion equation: range = u * t, where u is the initial horizontal velocity and range is the horizontal displacement.
The initial horizontal velocity is 15 m/s (since the ball is thrown horizontally). Plugging in the value of t (found in part a), we get:
range = 15 * 3.78
range ≈ 56.7 m
To find the angle, we can use the trigonometric relationship: tan(angle) = vertical displacement / horizontal displacement.
The vertical displacement is -70 m (since the ball falls downward), and the horizontal displacement is 56.7 m (range). Plugging in these values into the trigonometric relationship, we get:
tan(angle) = -70 / 56.7
Taking the inverse tangent (tan^-1) of both sides, we find:
angle = tan^-1(-70 / 56.7)
angle ≈ -51.4°
Therefore, the angle at which the ball hits the water is approximately 51.4° below the horizontal.
c) The speed at which it hits the water:
To find the speed, we can use the horizontal and vertical components of the velocity. Since the ball was thrown horizontally, the initial vertical velocity is zero. The initial horizontal velocity is 15 m/s (given).
Using the Pythagorean theorem, we can find the magnitude of the velocity:
v = √(u^2 + v^2)
Plugging in the values, we get:
v = √(15^2 + 0^2)
v = √225
v = 15 m/s
Therefore, the speed at which the ball hits the water is 15 m/s.