Since the equation for time=sqrt(2h/g)
set 2 seconds for time and 9.8 for gravity
so 2=sqrt(2h/9.8)
clear the sqrt by squaring both sides
4= 2h/9.8
9.8*4 =2h
(9.8*4)/2 = height.
now that you have the height, you can do v=distance/time
v=height from the equation prior/2 seconds
i hope that works..
The acceleration of the rock is calculated by dividing the change in velocity by the time taken. In this case, the change in velocity is 4.9 m/s and the time taken is 3 seconds. Thus, the acceleration of the rock is 4.9 m/s^2.
The velocity of the rock as it reaches the ground after 3.5 seconds of free fall can be calculated using the equation v = gt, where v is the final velocity, g is the acceleration due to gravity (approximately 9.81 m/s^2), and t is the time in seconds. Substituting the values, v = 9.81 m/s^2 * 3.5 s = 34.335 m/s. So, the velocity of the rock as it reaches the ground is approximately 34.34 m/s.
Assuming the rock is dropped (initial velocity is 0), the time it takes for the rock to reach the ground can be calculated using the formula: time = sqrt(2 * distance / gravity). The acceleration due to gravity is approximately 32 ft/s^2. Plugging in the values, the time it takes for the rock to reach the ground would be approximately 3 seconds.
The acceleration of the rock would be (1.63 , \text{m/s}^2) (calculated by dividing the change in velocity by the time taken).
The speed of the rock after 5 seconds, neglecting air resistance, can be determined using the equation: speed = initial velocity + acceleration * time. Since the initial speed is 0 m/s, and assuming the acceleration due to gravity is 9.81 m/s², the speed of the rock after 5 seconds would be 49.05 m/s.
176.4 meters
49
speed is decided by v = a(t), where t is your time variable and a is your acceleration constant. after 3 seconds, your velocity v = (9.8m/s2)(3s) = 29.4m/s
29.4
49
The acceleration of the rock is calculated by dividing the change in velocity by the time taken. In this case, the change in velocity is 4.9 m/s and the time taken is 3 seconds. Thus, the acceleration of the rock is 4.9 m/s^2.
S=ut+0.5*a*t2 s=displacement u=intial velocity=0m/s t=time=7seconds a=9.81m/s2 s=0*7+0.5*9.81*72 s=240.345m
The velocity of the rock as it reaches the ground after 3.5 seconds of free fall can be calculated using the equation v = gt, where v is the final velocity, g is the acceleration due to gravity (approximately 9.81 m/s^2), and t is the time in seconds. Substituting the values, v = 9.81 m/s^2 * 3.5 s = 34.335 m/s. So, the velocity of the rock as it reaches the ground is approximately 34.34 m/s.
,033
Assuming the rock is dropped (initial velocity is 0), the time it takes for the rock to reach the ground can be calculated using the formula: time = sqrt(2 * distance / gravity). The acceleration due to gravity is approximately 32 ft/s^2. Plugging in the values, the time it takes for the rock to reach the ground would be approximately 3 seconds.
The acceleration of the rock would be (1.63 , \text{m/s}^2) (calculated by dividing the change in velocity by the time taken).
Assuming you throw the rock horizontally off the cliff it drops down at the acceletrtion of gravity. height= 1/2 gt^2 With g = 9.8 m/sec and t = 5 seconds we have height = (1/2) (9.8)(5)(5) = 122.5 meters notice it has nothing todo with the 50 meter distance, which depends on the horizontal velocity.