When chasing a hare along a flat stretch of ground, a greyhound leaps into the air at a speed of 10.1 m/s, at an angle of 16.9°
above the horizontal. (a) What is the range of his leap? (b) For how much time is he in the air?
1 answer:
Answer:
a) The range of the leap is 5.80 m.
b) The time that the greyhound is in the air is 0.600 s
Explanation:
The position of the greyhound is given by the position vector "r":
r =(x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector at time "t".
x0 = initial horizontal position.
v0 = initial horizontal velocity.
y0 = initial vertical position
t = time.
α = leap angle.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
a) At the end of the leap, the position vector will be "r final" (see figure). The vector r final could be written as:
r final = (x, 0 m)
Where "x" is the range.
Then, using the equation for the vertical position we can obtain the time of flight of the leap. With that time, we can calculate the range using the equation for the horizontal position.
Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
0 m = 0 m + 10.1 m/s · t · sin 16.9° - 1/2 · 9.8 m/s² · t²
0 m = t (10.1 m/s · sin 16.9° - 4.9 m/s² · t)
0 m = 10.1 m/s · sin 16.9° - 4.9 m/s² · t
- 10.1 m/s · sin 16.9° / - 4.9 m/s² = t
t = 0.600 s
The x-component of the vector "r final" will be:
x = x0 + v0 · t · cos α
x = 0 m + 10.1 m/s · 0.600 s · cos 16.9°
x = 5.80 m
The range of the leap is 5.80 m.
b) The time that the greyhound is in the air is 0.600 s. It was calculated above using the equation for the y-component of the vector "r final".
You might be interested in
Answer:
The current through it will also decrease to half of its former value because according to Ohm's law the current flowing through a resistor is directly proportional to the potential difference applied across its ends provided that the temperature and some other necessary conditions remain constant.
This is mathematically represented as follows;
The current is thus given as
if R is constant and V is reduced to half, then we have the following;
Simplifying further we obtain
Equation (3) shows that the current I is also reduced to half.
Answer:
a) t=1s
y = 10.1m
v=5.2m/s
b) t=1.5s
y =11.475 m
v=0.3m/s
c) t=2s
y =10.4 m
v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Explanation:
Conceptual analysis
We apply the free fall formula for position (y) and speed (v) at any time (t).
As gravity opposes movement the sign in the equations is negative.:
y = vi*t - ½ g*t2 Equation 1
v=vit-g*t Equation 2
y: The vertical distance the ball moves at time t
vi: Initial speed
g= acceleration due to gravity
v= Speed the ball moves at time t
Known information
We know the following data:
Vi=15 m / s
t=1s ,1.5s,2s
Development of problem
We replace t in the equations (1) and (2)
a) t=1s
=15-4.9=10.1m
v=15-9.8*1 =15-9.8 =5.2m/s
b) t=1.5s
=22.5-11.025=11.475 m
v=15-9.8*1.5 =15-14.7=0.3m/s
c) t=2s
= 30-19.6=10.4 m
v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)