An object is thrown up from the top of building 50 m high. it rises to a maximum height of 20 m above the roof.?
a) when does it land on the ground
b) at what velocity does it land?
c) what is its velocity 15m below the roof?
a. h = 50 + 20 = 70 m Above gnd.
V^2 = Vo^2 + g*h = 0
Vo^2 - 9.8*20 = 0
Vo^2 = 196
Vo = 14 m/s. = Initial velocity.
V = Vo + g*Tr = 0
14 - 9.8Tr = 0
9.8Tr = 14
Tr = 1.43 s. = Rise time.
h = 0.5g*t^2 = 70 m.
4.9t^2 = 70
t^2 = 14.3
Tf = 3.78 s. = Fall time.
Tr+Tf = 1.43 + 3.78 = 5.21 s. to reach
gnd.
b. V = Vo + g*Tf
Vo = 0
g = 9.8 m/s^2
Tf = 3.78 s.
Solve for V.
c. V^2 = Vo^2 + 2g*h
Vo = 0
g = 9.8 m/s^2
h = 70-35 = 35 m.
Solve for V.
To answer these questions, we first need to determine the time it takes for the object to reach its maximum height.
We can use the kinematic equation for displacement to find the time it takes for the object to reach its maximum height:
\[s = u \cdot t + \frac{1}{2} \cdot a \cdot t^2\]
Since the object is thrown upwards, its initial velocity (u) is positive and the acceleration due to gravity (a) is negative (taking downwards as the negative direction). The displacement (s) is the difference between the initial and final heights, which is 20 m - 50 m = -30 m.
Therefore, the equation becomes:
\[-30 = u \cdot t + \frac{1}{2} \cdot (-9.8) \cdot t^2\]
Simplifying the equation, we have:
\[-30 = u \cdot t - 4.9 \cdot t^2\]
Since the initial velocity is not given, we need additional information to proceed.