Madhuri throws a stone the stone is thrown horizontally at 33.9 m/s from the top of a 297.9m cliff.
1.How high is the stone after 3.35 sec?
2.What is the range?
equations to find please.
h=297.9m-4.9t
distance=33.9*t
0.5*9.8*(3.35)^2
To solve this problem, we can use the equations of motion for projectiles. The two main equations we will use are:
1. The vertical displacement equation: h = h0 + v0y * t + (1/2) * g * t^2
2. The horizontal displacement equation: R = v0x * t
Where:
- h is the vertical displacement of the stone.
- h0 is the initial height of the stone (in this case, the height of the cliff).
- v0y is the vertical component of the initial velocity.
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
- t is the time.
- R is the horizontal displacement of the stone.
- v0x is the horizontal component of the initial velocity.
Let's solve each part of the problem using these equations:
1. How high is the stone after 3.35 seconds?
In this case, we can assume that the stone is thrown horizontally, so the initial vertical velocity (v0y) is zero. Therefore, the vertical displacement equation simplifies to:
h = h0 + (1/2) * g * t^2
Substituting the given values:
h = 297.9 m + (1/2) * 9.8 m/s^2 * (3.35 s)^2
h ≈ 297.9 m + 51.49 m
h ≈ 349.39 m
Therefore, the stone is approximately 349.39 meters high after 3.35 seconds.
2. What is the range?
The range is the horizontal displacement of the stone. In this case, we need to find the horizontal component of the initial velocity (v0x). Since the stone is thrown horizontally, the initial horizontal velocity is the same as the magnitude of the total initial velocity (33.9 m/s).
Using the horizontal displacement equation, we get:
R = v0x * t
R = 33.9 m/s * 3.35 s
R ≈ 113.47 m
Therefore, the range of the stone is approximately 113.47 meters.