An outfielder throws a baseball to home plate with a velocity of +15 m/s and an angle of 30°. When will the ball reach its highest altitude?
(Points : 3)
1.2 s
1.5 s
0.77 s
0.50 s
The vertical component of the balls velocity is the following:
15m/s*Sin 30=7.5m/s
The deceleration due to gravity is 9.8m/s. So, after 1s the ball will be falling back down to the ground, which eliminates the first two answer choices. Does it take a half a second for the ball's velocity to reach 0m/s, or does it take slightly more time for the ball's velocity to reach 0m/s? In half a second, the ball will decelerate by 4.8m/s, which means that it will still have a velocity of 2.7m/s. The best answer choice is 0.77s.
****You could have just used an equation, but understanding it conceptually, is better at times.
If you want the equation, here you go:
Vf=Vi+gt
Where
Vf=0m/s
Vi=7.5m/s
g=-9.8m/s^2
and
t=??
0=7.5m/s-9.8m/s^2*t
-7.5m/s/-9.8m/s^2=t
t=0.765
Vi = 15 sin 30 = 7.5
v = Vi - gt
v = 0 at top
so
g t = 7.5
t = 7.5/9.8 = about .77 s
Well, if the outfielder throws a baseball, I hope they don't throw it too high or it might end up hitting a cloud! But to answer your question, the ball will reach its highest altitude at 0.77 seconds. It's like a brief moment of weightlessness for the baseball before gravity brings it back down to earth.
To find the time at which the ball reaches its highest altitude, we can use the kinematic equation for vertical motion:
y = y0 + v0y*t - (1/2)*g*t^2
where:
y = vertical position (altitude)
y0 = initial vertical position (assumed to be zero)
v0y = vertical component of initial velocity
g = acceleration due to gravity (assumed to be 9.8 m/s^2)
t = time
At the highest point, the vertical velocity will be zero, so we can set v0y - g*t = 0 and solve for t.
Given:
v0y = 15 m/s * sin(30°) ≈ 7.5 m/s
Setting the equation v0y - g*t = 0 and solving for t, we get:
7.5 m/s - 9.8 m/s^2 * t = 0
Solving for t:
t = 7.5 m/s / 9.8 m/s^2
t ≈ 0.77 s
Therefore, the ball will reach its highest altitude at approximately 0.77 seconds. So the correct answer is 0.77 s.
To determine when the ball will reach its highest altitude, we can use the concept of projectile motion. The formula to find the time at which the projectile reaches its highest point is given by the equation:
t = vy / g
where vy is the vertical component of velocity and g is the acceleration due to gravity (approximately 9.8 m/s²).
In this case, the ball was thrown with a velocity of +15 m/s at an angle of 30°. To find the vertical component of velocity, we multiply the initial velocity (15 m/s) by the sine of the angle (30°):
vy = 15 m/s * sin(30°)
vy ≈ 7.5 m/s
Plugging this value into the formula, we have:
t = 7.5 m/s / 9.8 m/s²
t ≈ 0.77 s
Therefore, the ball will reach its highest altitude approximately 0.77 seconds after it was thrown.
Hence, the correct answer is: 0.77 s.