Question:
an object is thrown upward with the velocity of 25m/s then find the maximum height reached by object also draw a V-T graph for object till it reaches the ground take gravitational acceleration be 10 m/s²
Answer:
To find the maximum height reached by the object, we can use the kinematic equation for vertical motion:
v2 = u2 + 2as
Where:v = final velocity (0 m/s at the maximum height)
u = initial velocity (25 m/s)
a = acceleration due to gravity (-10 m/s^2, negative since it acts opposite to the motion)
s = displacement (maximum height)
Plugging in the values, we get:
0^2 = (25 m/s)2 + 2×(-10 m/s2)×s
Simplifying the equation:
0 = (625 m2/s2 )- (20 m/s2 × s)
20 m/s2× s = 625 m2/s2
s = 625 m2/s2 / 20 m/s2
s = 31.25 meters
Therefore, the maximum height reached by the object is 31.25 meters.
To draw a V-T (velocity-time) graph for the object until it reaches the ground, we need to consider the following:
1. The object is thrown upward, so its initial velocity is 25 m/s in the upward direction.
2. The acceleration due to gravity is acting in the downward direction and is constant at -10 m/s2.
3. At the maximum height, the velocity of the object becomes 0 m/s.
4. The object will continue to move upward until its velocity becomes 0 m/s, and then it will start falling back down.
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