Two objects were
projected vertically
upwards at different
time with a speed ofTwo objects were
projected vertically
upwards at different
time with a speed of
80ms-¹ and 100ms-¹
respectively.if the time
interval is 2 sec,when
and where would they
meet.? 65,485 results
80ms-¹ and 100ms-¹
respectively.if the time
interval is 2 sec,when
and where would they
meet.? 65,485 results
Two objects were
projected vertically
upwards at different
time with a speed ofTwo objects were
projected vertically
upwards at different
time with a speed of
80ms-¹ and 100ms-¹
respectively.if the time
interval is 2 sec,when
and where would they
meet.?
Two objects were
projected vertically
upwards at different
time with a speed ofTwo objects were
projected vertically
upwards at different
time with a speed of
80ms-¹ and 100ms-¹
respectively.if the time
interval is 2 sec,when
and where would they
meet.? a
geez - what do you have against just typing your stuff normally?
You want to find t where
80t - 4.9t^2 = 100(t-2) - 4.9(t-2)^2
then use that to find the height
To find out when and where the two objects would meet, we can use the kinematic equations for motion in one dimension (vertical direction).
Let's assume that the object with an initial speed of 80 m/s was thrown first, and the object with an initial speed of 100 m/s was thrown 2 seconds later.
First, let's find the time it takes for the first object to reach the meeting point. We can use the kinematic equation:
y = v₀t + (1/2)gt²
where:
- y is the displacement (height above the starting point),
- v₀ is the initial vertical velocity,
- t is the time, and
- g is the acceleration due to gravity (approximately 9.8 m/s²).
For the first object:
y = 0 (since it starts at the same height as the meeting point)
v₀ = 80 m/s
t = t₁ (time taken by the first object)
0 = 80t₁ - (1/2)(9.8)(t₁)²
Simplifying the equation:
0 = 80t₁ - 4.9t₁²
Now, let's find the time it takes for the second object to reach the meeting point. Since it is thrown 2 seconds after the first object, its time would be t₁ + 2.
For the second object:
y = 0 (since it starts at the same height as the meeting point)
v₀ = 100 m/s
t = t₁ + 2 (since it is thrown 2 seconds after the first object)
0 = 100(t₁ + 2) - (1/2)(9.8)(t₁ + 2)²
Simplifying the equation:
0 = 100(t₁ + 2) - 4.9(t₁ + 2)²
Now, we have two equations with two unknowns (t₁ and t₂). We can solve these equations simultaneously to find the values of t₁ and t₂.
Once we have the values of t₁ and t₂, we can substitute them into either of the two equations to find the height (y) at which they meet.
It is important to note that solving these equations may require numerical methods or trial and error, as they are nonlinear equations.