use F = ma
F : force m : mass a : acceleration
so
f = 5kg * 20 m/s2 = 100 N
The tennis ball lands at a point 40.4 m from the base of the building.
The tennis ball is projected with a horizontal velocity <em>u</em> from a window, which is at a height <em>y</em> from the ground. The ball lands at a distance <em>x</em> from the base of the building. Let the ball take a time <em>t</em> to reach the ground. In the time <em>t</em> ,the ball falls a vertical distance <em>y</em> and also travel a horizontal distance <em>x</em>.
The initial vertical velocity of the ball is zero, since the ball is projected in the horizontal direction. The ball falls down under the action of gravitational force.
Thus, use the equation of motion,
rewrite the expression for <em>t</em> and calculate the value of <em>t</em> using 9.81 m/s²for <em>g</em> and 500 m for <em>y</em>.
The horizontal distance <em>x</em> is traveled using the constant velocity <em>u </em>since no force acts on the ball in the horizontal direction.
Therefore,
Substitute 4 m/s for <em>u</em> and 10.096 s for <em>t</em>
Thus, the ball lands at a point 40.4 m from the base of the building.
When the grasshoppers vertical velocity is exactly zero.
v = -g•t + v0.
v: vertical part of velocity. Is zero at maximum height.
g: 9.81
t: time you are looking for
v0: initial vertical velocity
Find the vertical part of the initial velocity, by using the angle at which the grasshopper jumps.
Answer:
The angle that will have the most range (Goes further) <u>will be the one that is shot with 60 ° with the horizontal. </u>
Since the one that shoots with respect to the vertical, it will only reach higher. And it won't go as far as the horizontal.
Greetings.
Danna.
Answer:
1830 m /s
Explanation:
The potential energy of the projectile at the surface of planet
- G x 1.46 x 10²³ m / 5 x 10⁶
Kinetic energy at surface
1/2 m v²
Total energy = - G x 1.46 x 10²³ m / 5 x 10⁶ + 1/2 m v²
- 6.67 X 10⁻¹¹ x 1.46 x 10²³ /5 x 10⁶ m + 1/2 m x 2000²
- 1.947 x 10⁶ m + 2m x 10⁶
= .053 x 10⁶ m
Potential energy of projectile at 1000 km height
= - G x 1.46 x 10²³ m / 6 x 10⁶
= - 1.6225 x 10⁶ m J
Total energy
= - 1.6225 x 10⁶ + 1/2 m V ²
Applying conservation of energy in gravitational field at surface and height
- 1.6225 x 10⁶ m + 1/2 m V ² = .053 x 10⁶ m
1/2 V ² = ( .053 + 1.6225 ) x 10⁶
1/2 V² = 1.6755 x 10⁶ m /s
V = 1.83 X 10³ m/s
1830 m /s
