A ball is thrown horizontally at 15 m/s from a clifftop 60 m above the sea level. Find:
a) the time to land;
b) the angle at which it hits the water;
° below the horizontal
c) the speed at which it hits the water.
m/s
To find the answers, we can use the following equations of motion:
1) Vertical motion equation:
h = ut + 0.5gt^2
2) Horizontal motion equation:
d = vt
where:
h = height (60 m)
u = initial vertical velocity (0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time to land
d = horizontal distance
v = horizontal velocity (15 m/s)
Now let's solve for each part of the problem step by step:
a) Time to Land:
Using the vertical motion equation, we can set h = 0 and solve for t:
0 = (0) t + 0.5(-9.8) t^2
0 = -4.9t^2
Solving for t gives us:
t = 0 or t = √(0/-4.9)
Since time cannot be negative, t = 0 is not considered. Therefore, the ball lands after t = √(0/-4.9) seconds.
b) Angle at which it hits the water:
To find this angle, we need to find the vertical velocity component of the ball at the time of impact. The horizontal velocity component remains constant throughout the motion.
Using the vertical motion equation, we can find the final vertical velocity (v_f) at h = 0:
0 = (0) + (-9.8)(t)
0 = -9.8t
t = 0 or t = √(0/-9.8)
Since time cannot be negative, t = 0 is not considered. Therefore, the vertical velocity component at the time of impact is v_f = √(0/-9.8) m/s.
To find the angle below the horizontal, we can use the following trigonometric relationship:
tan(θ) = v_f / v
tan(θ) = √(0/-9.8) / 15
θ = tan^(-1)(√(0/-9.8) / 15)°
c) Speed at which it hits the water:
To find the speed at which it hits the water, we can use the horizontal motion equation:
d = vt
Substituting the given values, we can solve for d:
d = 15 * √(0/-4.9) m
The speed of the ball hitting the water is equal to the magnitude of the horizontal velocity component, so it is 15 m/s.
To find the answers to these questions, we can use the principles of projectile motion. When an object is thrown horizontally, its vertical motion can be treated independently from its horizontal motion.
a) The time to land can be found by using the equation for vertical displacement:
\[h = \frac{1}{2}gt^2\]
where h is the vertical displacement (in this case, 60 meters) and g is the acceleration due to gravity (approximately 9.8 m/s^2). Plugging in these values, we get:
\[60 = \frac{1}{2} \times 9.8 \times t^2\]
Simplifying the equation, we have:
\[t^2 = \frac{60 \times 2}{9.8}\]
\[t^2 \approx 12.24\]
Taking the square root of both sides, we find:
\[t \approx 3.50 seconds\]
So, the ball will take approximately 3.50 seconds to land.
b) The angle at which the ball hits the water can be determined by calculating the time it takes for the ball to reach the water horizontally. The horizontal displacement of the ball can be found using:
\[d = v \times t\]
where d is the horizontal displacement and v is the initial horizontal velocity (which is equal to 15 m/s in this case). Since the horizontal displacement is the same as the initial velocity multiplied by the time, we have:
\[d = 15 \times 3.50\]
\[d \approx 52.5 meters\]
The vertical displacement remains the same (60 meters) as given in the problem. To find the angle, we can use the tangent of the angle:
\[tan(\theta) = \frac{60}{52.5}\]
Simplifying the equation, we have:
\[\theta \approx tan^{-1}(1.143)\]
\[\theta \approx 48.53^\circ\]
So, the angle at which the ball hits the water is approximately 48.53° below the horizontal.
c) The speed at which the ball hits the water can be found using the Pythagorean theorem. The horizontal and vertical components of the velocity can be calculated as follows:
\[v_x = v \cdot cos(\theta)\]
\[v_y = v \cdot sin(\theta)\]
where v is the initial velocity (15 m/s) and \(\theta\) is the angle at which it hits the water (48.53°). Plugging in the values, we have:
\[v_x = 15 \cdot cos(48.53)\]
\[v_x \approx 9.789\]
\[v_y = 15 \cdot sin(48.53)\]
\[v_y \approx 11.362\]
The total velocity can be found using the Pythagorean theorem:
\[v_{total} = \sqrt{v_x^2 + v_y^2}\]
\[v_{total} = \sqrt{9.789^2 + 11.362^2}\]
\[v_{total} \approx 14.72 m/s\]
So, the speed at which the ball hits the water is approximately 14.72 m/s.