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 the given problem, we can use the equations of motion for free fall and projectile motion. Let's break down each part of the problem step by step:
a) To find the time it takes for the ball to land, we need to calculate the time it takes for the ball to fall vertically from the clifftop to the water surface below. We can use the equation for the time of flight in free fall:
t = sqrt((2 * h) / g)
Where:
t = time of flight
h = height of the cliff = 70 m
g = acceleration due to gravity = 9.8 m/s^2 (assuming no air resistance)
Plugging the given values into the equation, we have:
t = sqrt((2 * 70) / 9.8)
t ≈ sqrt(14.286)
t ≈ 3.78 s
Therefore, it takes approximately 3.78 seconds for the ball to land.
b) To find the angle at which the ball hits the water, we can use the horizontal and vertical components of the ball's velocity. Since the horizontal velocity remains constant throughout the motion, the angle can be determined from the vertical component of velocity. The angle can be found using the formula:
tan(θ) = (v_y / v_x)
Where:
θ = the angle with the horizontal
v_y = vertical component of velocity
v_x = horizontal component of velocity
In this case, the horizontal component of velocity is given as 15 m/s, and the vertical component of velocity is determined using the equation of motion for free fall:
v_y = g * t
Using the previously calculated time (t ≈ 3.78 s), the vertical component of velocity can be found:
v_y = 9.8 * 3.78
v_y ≈ 37.044 m/s
Now, we can calculate the angle using the formula:
θ = atan(v_y / v_x)
θ = atan(37.044 / 15)
θ ≈ 68.22°
Therefore, the ball hits the water at an angle of approximately 68.22° below the horizontal.
c) Finally, to find the speed at which the ball hits the water, we can use the Pythagorean theorem to calculate the resultant velocity. The resultant velocity is given by:
v = sqrt(v_x^2 + v_y^2)
Using the horizontal and vertical components of velocity obtained earlier:
v = sqrt(15^2 + 37.044^2)
v ≈ sqrt(225 + 1373.154)
v ≈ sqrt(1598.154)
v ≈ 39.98 m/s
Therefore, the ball hits the water with a speed of approximately 39.98 m/s.