distance = 1/2*g*t^2
solve for t. Check m thinking.
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18 m/s. the cliff is 50 m above a flat, horizontally beach. how long after being released does the stone strike the beach below the cliff?
2 answers
We can talk about this projection as two simultaneous and perpendicular motion.
Assuming that the height of the student can be omitted, the gravity is a constant with a value of 10m/s, no friction or disturbance in air and the Earth is an inertial frame:
The time of flight can be gained by using S = ut + (1/2)*at^2 downwards and as the initial value for the vertical velocity is 0 we can omit the 'ut' part.
50 = (1/2)*10t^2
t = √10 s
The horizontal distance can be gained through S = ut (for the rock flies with a uniform horizontal velocity and with no horizontal acceleration).
S = 18*√10
S = 18√10 m
Assuming that the height of the student can be omitted, the gravity is a constant with a value of 10m/s, no friction or disturbance in air and the Earth is an inertial frame:
The time of flight can be gained by using S = ut + (1/2)*at^2 downwards and as the initial value for the vertical velocity is 0 we can omit the 'ut' part.
50 = (1/2)*10t^2
t = √10 s
The horizontal distance can be gained through S = ut (for the rock flies with a uniform horizontal velocity and with no horizontal acceleration).
S = 18*√10
S = 18√10 m
1 answer