A tennis ball is thrown horizontally from an elevation of 20.60 m above the ground with a speed of 21.0 m/s
vertical m from ground???
f the ball is still in the air, how long before it hits the ground? (If the ball is not in the air, enter 0.)????
If the ball is still in the air, where will it be with respect to the starting point once it lands???
Please helppp
a. h = Vo*t + 0.5g*t^2 = 20.6 m.
0 + 4.9t^2 = 20.6,
t^2 = 4.20,
Tf = 2.05 s. = Time to fall to ground.
b. Dx = Xo * Tf,
Dx = 21m/s * 2.05s = 43.05 m.
To find the vertical distance from the ground, we can use the equation of motion:
h = v₀t + (1/2)gt²
where:
h - vertical distance from the ground (to be determined)
v₀ - initial vertical velocity (0 m/s since the ball is thrown horizontally)
t - time
g - acceleration due to gravity (approximately 9.8 m/s²)
Since the initial vertical velocity is zero, the equation simplifies to:
h = (1/2)gt²
Substituting the values:
h = (1/2)(9.8 m/s²)(t²)
Now, let's calculate the vertical distance from the ground.
h = (1/2)(9.8 m/s²)(t²)
h = 4.9t²
Since we don't know the time at which the ball hits the ground, we cannot determine the vertical distance using only the given information.
To find the time before the ball hits the ground, we can use the equation of motion for vertical motion:
h = v₀t + (1/2)gt²
Since the ball is thrown horizontally, the vertical initial velocity is zero. The equation simplifies to:
h = (1/2)gt²
Plugging in the values:
20.60 m = (1/2)(9.8 m/s²)t²
41.20 m = 9.8 m/s²t²
t² = 41.20 m / (9.8 m/s²)
t² = 4.20 s²
t = √(4.20 s²)
t ≈ 2.05 s
Therefore, it will take approximately 2.05 seconds for the ball to hit the ground.
Since the ball is thrown horizontally, it will be at the same horizontal position as the starting point once it lands.