m = mass of the cart = 85 kg
F = net force on the cart = 250 N
a = acceleration of the cart
acceleration of the cart is given as
a = F/m
a = 250/85
a = 2.94 m/s²
t = time for which the force is applied = 5 sec
v₀ = initial velocity of the cart = 0 m/s
v = final velocity of the cart just before it flies off the cliff = ?
using the equation
v = v₀ + a t
inserting the values
v = 0 + (2.94) (5)
v = 14.7 m/s
consider the motion of cart after it flies off the cliff in vertical direction :
v' = initial velocity in vertical direction = 0 m/s
a' = acceleration in vertical direction = g = acceleration due to gravity = 9.8 m/s²
t' = time taken for the cart to land = ?
Y' = vertical displacement of the cart = height of cliff = 100 m
using the kinematics equation
Y' = v' t' + (0.5) a' t'²
100 = (0) t' + (0.5) (9.8) t'²
t' = 4.52 sec
consider the motion of cart along the horizontal direction after it flies off the cliff
X = distance traveled from the base of cliff = ?
t' = time of travel = 4.52 sec
v = velocity along the horizontal direction = 14.7 m/s
distance traveled from the base of cliff is given as
X = v t'
X = 14.7 x 4.52
X = 66.4 m
