A placekicker kicks a football with a velocity of 35 m/sec and an angle of 25˚
above the horizontal. When the football is 10 m off the ground,
a. What is the horizontal velocity of the ball?
b. What is the vertical velocity of the ball?
c. What is the total velocity of the ball?
d. How long did it take for the ball to reach that point (10 m above the
ground)?
To answer these questions, we can use the principles of projectile motion. Projectile motion is the motion of an object that is launched into the air and moves along a curved path under the influence of gravity.
a. The horizontal velocity of the ball remains constant throughout its motion because there is no horizontal acceleration. We can find the horizontal velocity using the given information. The velocity is given as 35 m/sec, and the angle is given as 25˚ above the horizontal. We can use trigonometry to find the horizontal velocity.
The horizontal velocity is given by the formula: horizontal velocity = velocity * cos(angle)
In this case, the horizontal velocity = 35 m/sec * cos(25˚).
b. The vertical velocity of the ball changes due to the effect of gravity. We can find the vertical velocity using the given information. The velocity is given as 35 m/sec, and the angle is given as 25˚ above the horizontal.
The vertical velocity is given by the formula: vertical velocity = velocity * sin(angle)
In this case, the vertical velocity = 35 m/sec * sin(25˚).
c. The total velocity of the ball is the combination of the horizontal and vertical velocities. We can use the Pythagorean theorem to find the total velocity.
The total velocity is given by the formula: total velocity = sqrt(horizontal velocity^2 + vertical velocity^2)
In this case, the total velocity = sqrt((35 m/sec * cos(25˚))^2 + (35 m/sec * sin(25˚))^2).
d. To find the time taken for the ball to reach a certain height, we can use the equations of motion. We know that the vertical motion follows a parabolic path due to gravity, and the time taken to reach a specific height can be found using the equation:
height = initial vertical velocity * time + (1/2) * acceleration * time^2
In this case, the height is given as 10 m, the initial vertical velocity can be found using the vertical velocity calculated in part b, and the acceleration is -9.8 m/sec^2 (due to gravity). We can solve this equation to find the time taken for the ball to reach a height of 10 m.
Vo = 35m/s[25o]. = Initial velocity.
Xo = 35*Cos25 = 31.7 m/s = Hor. component of initial velocity.
Yo = 35*sin25 = 14.8 m/s = Ver. component of initial velocity.
a. X = Xo = 31.7 m/s and does not change.
b. Y^2 = Yo^2 + 2g*h = (14.8)^2 -19.6*10 = 23.04
Y = 4.8 m/s.
c. V = Xo + Yi = 31.7 + 4.8i = 32.1m/s[8.61o]
d. Y = Yo+g*t = 4.8 @ 10 m.
14.8 - 9.8t = 4.8
t = ?.
a) Vh = 35 m/s * cos(25º)
... this remains constant for the flight
c) 1/2 m v^2 = 1/2 m Vo^2 - (m g h)
... v^2 = Vo^2 - (2 g h) ... h = 10
b) Vv^2 = v^2 - Vh^2
d) h = -1/2 g t^2 + Vo sin(25º) t
... 10 = -4.9 t^2 + 35 sin(25º) t
solve for t ... the smaller value is going up, the larger is coming down