A stone is thrown vertically upward with [tex]$40 m / s$[/tex]. Calculate:
(i) velocity of the stone after 1.5 sec (Use [tex]$g =9.8 m / s ^2$[/tex])
(ii) time taken to reach max height
(iii) total distance travelled by the stone before return back to the surface.
Answer (1)
Calculate the velocity after 1.5 seconds using v = u − g t , resulting in v = 40 − 9.8 \tims 1.5 = 25.3 m / s .
Determine the time to reach maximum height by setting v = 0 in the same equation, giving t = a rc 40 9.8 a pp ro x 4.0816 s .
Compute the maximum height using v 2 = u 2 − 2 g s , which yields s = a rc 4 0 2 2 \tims 9.8 a pp ro x 81.6327 m .
Find the total distance travelled by doubling the maximum height, resulting in a total distance of approximately 163.2653 m .
Explanation
Problem Setup We are given that a stone is thrown vertically upward with an initial velocity of 40 m / s . We need to calculate (i) the velocity of the stone after 1.5 seconds, (ii) the time taken to reach maximum height, and (iii) the total distance travelled by the stone before returning to the surface. We will use the equations of motion to solve this problem.
Velocity after 1.5 seconds (i) To find the velocity of the stone after 1.5 seconds, we use the equation: v = u − g t where:
v is the final velocity,
u is the initial velocity ( 40 m / s ),
g is the acceleration due to gravity ( 9.8 m / s 2 ),
t is the time ( 1.5 s ).
Substituting the given values, we get: v = 40 − ( 9.8 × 1.5 ) = 40 − 14.7 = 25.3 m / s So, the velocity of the stone after 1.5 seconds is 25.3 m / s .
Time to reach maximum height (ii) To find the time taken to reach maximum height, we use the fact that the final velocity at the maximum height is 0. Using the same equation: v = u − g t At maximum height, v = 0 , so: 0 = 40 − 9.8 t Solving for t , we get: t = 9.8 40 ≈ 4.0816 s So, the time taken to reach maximum height is approximately 4.0816 seconds.
Total distance travelled (iii) To find the total distance travelled, we first find the maximum height reached by the stone using the equation: v 2 = u 2 − 2 g s At maximum height, v = 0 , so: 0 = 4 0 2 − 2 × 9.8 × s Solving for s , we get: s = 2 × 9.8 4 0 2 = 19.6 1600 ≈ 81.6327 m The total distance travelled is twice the maximum height, since the stone travels up and then down the same distance: T o t a l D i s t an ce = 2 × s = 2 × 81.6327 ≈ 163.2653 m So, the total distance travelled by the stone is approximately 163.2653 meters.
Final Answer Therefore, the velocity of the stone after 1.5 seconds is 25.3 m / s , the time taken to reach maximum height is approximately 4.0816 s , and the total distance travelled by the stone is approximately 163.2653 m .
Examples
Understanding projectile motion is crucial in various real-world scenarios. For instance, engineers use these principles to design ballistics for launching satellites into orbit or to calculate the trajectory of projectiles in games. Athletes also use these concepts intuitively when throwing a ball or hitting a golf ball, optimizing the angle and velocity to achieve maximum distance or accuracy. Even in weather forecasting, understanding the vertical motion of air parcels helps predict cloud formation and precipitation.
